BY THE SAME AUTHOR
WAR SHIPS
A Textbook on the Construction, Protection,
Stability, Turning, etc., of War Vessels
With 219 Diagrams and 48 pages of plain paper
at end for Notes and References
Medium 8vo, I2S. 6d. net
A TEXTBOOK ON LAYING OFF
or, the Geometry of Shipbuilding
By EDWARD L. ATTWOOD, M.Inst.N.A.,
R.C.N.C., and I. C. G. COOPER, Senior Lofts
man, H.M. Dockyard, Chatham, etc With
121 Diagrams. 8vo, 6s. net.
LONGMANS, GREEN AND CO.
LONDON, NEW YORK, BOMBAY, CALCUTTA AND MADRAS
TEXTBOOK
OF
THEORETICAL NAVAL
ARCHITECTURE ,..,
BY *' '*, :,, ; ...'K:'
EDWARD L. ATTWOOD, M. INST. N.A.
MEMBER OF ROYAL CORPS OF NAVAL CONSTRUCTORS
FORMERLY LECTURER ON NAVAL ARCHITECTURE AT THE ROYAL NAVAL COLLEGI
GREENWICH
AUTHOR OF "WAR SHIPS" AND "THE MODERN WAR SHIP"
JOINT AUTHOR OF " LAYING OFF, OR THE GEOMETRY OF SHIPBUILDING "
WITH NUMEROUS DIAGRAMS
NEW IMPRESSION
LONGMANS, GREEN AND CO,
39 PATERNOSTER ROW, LONDON
FOURTH AVENUE & S OTH STREET, NEW YORK
BOMBAY, CALCUTTA, AND MADRAS
I9I7
All rights reserved
Engineering
Library
BIBLIOGRAPHICAL NOTE
First Printed .... February r , 1899.
Second Edition .... November ^ 1900.
Third Edition .... October^ 1902.
Fourth Edition. . . . September \ 1905.
Edition .... December , 1908.
Edition . . , . February ', 1912.
Reprinted March> 1915.
Edition .... y?/^, 1916.
Impression . . . fttly, 1917.
PREFACE
THIS book has been prepared in order to provide students and
draughtsmen engaged in Shipbuilders' and Naval Architects
drawing offices with a textbook which should explain the
calculations which continually have to be performed. It is
intended, also, that the work, and more especially its later
portions, shall serve as a textbook for the theoretical portion
of the examinations of the Science and Art Department in
Naval Architecture. It has not been found possible to include
all the subjects given in the Honours portion of the syllabus,
such as advanced stability work, the rolling of ships, the vibra
tion of ships, etc. These subjects will be found fully treated
in one or other of the books given in the list on page 488.
A special feature of the book is the large number of
examples given in the text and at the ends of the chapters.
By means of these examples, the student is able to test his
grasp of the principles and processes given in the text. It is
hoped that these examples, many of which have been taken
from actual drawing office calculations, will form a valuable
feature of the book.
Particulars are given throughout the work and at the end
as to the books that should be consulted for fuller treatment of
the subjects dealt with.
In the Appendix are given the syllabus and specimen
questions of the examination in Naval Architecture conducted
840838 a 3
vi Preface.
by the Science and Art Department. These are given by the
permission of the Controller of His Majesty's Stationery Office.
I have to thank Mr. A. W. Johns, Instructor in Naval
Architecture at the Royal Naval College, Greenwich, for
reading through the proofs and for sundry suggestions. I also
wish to express my indebtedness to Sir W. H. White, K.C.B.,
F.R.S., Assistant Controller and Director of Naval Construction
of the Royal Navy, for the interest he has shown and the
encouragement he has given me during the progress of the
book.
E. L. ATTWOOD.
LONDON,
February^ 1899.
PREFACE TO THE NEW
EDITION.
IN the present edition the matter has been somewhat re
arranged and a number of additions made. Two new chapters
have been added, one on launching calculations, and one on
the turning of ships.
LONDON, 1916.
REMARKS ON EDUCATION IN
NAVAL ARCHITECTURE
FOR the bulk of those who study the subject of Naval Archi
tecture, the only instruction possible is obtained in evening
classes, and this must be supplemented by private study. The
institutions in which systematic instruction in day courses
is given are few in number, viz. (i) Armstrong College,
University of Durham, NewcastleonTyne ; (2) University
of Glasgow ; (3) University of Liverpool ; (4) Royal Naval
College, Greenwich ; and students who can obtain the advan
tage of this training are comparatively few in number. An
account of the course at Glasgow is to be found in a paper
before the I.N.A. in 1889 by the late Prof. Jenkins, and at the
Royal Naval College, in a paper before the I.N.A. in 1905
by the writer ; see also a paper by Professor Welch on the
scientific education of naval architects before N.E. Coast
Institution, 1909. There are scholarships to be obtained for
such higher education, particulars of which can be had by
application to the Glasgow, Liverpool, and Newcastle Colleges,
to the Secretary of the Admiralty, Whitehall, S.W., and to the
Secretary of the Institution of Naval Architects, Adelphi
Terrace, Strand. In these courses it is recognized that the
study of other subjects must proceed concurrently with that
of Naval Architecture.
The Naval Architect has to be responsible for the ship
as a complete design, and in this capacity should have some
familiarity with all that pertains to a ship. Thus he should
know something of Marine Engineering (especially of pro
pellers) ; of Electricity and Magnetism ; of armour, guns and
gunmountings in warships; of masts, rig, etc., in sailing
viii Remarks on Education in Naval Architecture.
vessels ; of the work of the stevedore in cargo vessels ; of
questions relating to the docking and undocking of ships ; of
appliances for loading and unloading of ships ; of the regula
tions of the Registration Societies and the Board of Trade
regarding structure, freeboard, and tonnage ; of appliances for
navigating, as well as having a thorough knowledge of the
practical work of the shipyard. In the early stages of a design,
the naval architect frequently has to proceed independently in
trying alternatives for the desired result, and it is not until the
design is somewhat matured that he can call in the assistance
of specialists in other departments. The naval architect
should, therefore, have an interest in everything connected
with the type of ship he has to deal with, and he will con
tinually be collecting data which may be of use to him in his
subsequent work.
For the average student of Naval Architecture, in addition
to the work he does and observes in the shipyard, mould loft,
and drawing office, it is necessary to attend evening classes in
Naval Architecture and other subjects. The apprentice should
systematically map out his time for this purpose. In the first
place, a good grounding should be obtained in mechanical
drawing and in elementary mathematics. Both of these sub
jects are now taught by admirable methods. The drawing
classes are usually primarily intended for Engineering students,
but this is no drawback, as it will familiarize the student with
drawings of engineering details which he will find of consider
able service to him in his subsequent work. Some institutions
very wisely do not allow students to take up the study of any
special subject, as Naval Architecture, until they have proved
themselves proficient in elementary drawing and mathematics.
The time thus spent is a most profitable investment.
The Board of Education now only hold examinations in
two stages, a " lower " and a "higher," see p. 446, but teachers
will probably divide the work between these stages, and
themselves hold examinations.
We will suppose, then, that a student starts definitely with
the lowest class in Naval Architecture. With this subject
he should also take up Elementary Applied Mechanics, and,
Remarks on Education in Naval Architecture, ix
if possible, some Mathematics. The next year may be devoted
to the Board of Education Lower Examination in Naval
Architecture, with a course in more advanced Applied
Mechanics, and a course in Magnetism and Electricity or
Chemistry would form a welcome relief. The next year may
be devoted to further study in Mathematics, Theoretical and
Applied Mechanics, Electricity and Magnetism. The next
year may be devoted to another class in Naval Architecture,
with more advanced Mathematics, including the Differential
and Integral Calculus. This latter branch of mathematics is
essential in order to make any progress in the higher branches
of any engineering subject. If the student is fortunate enough
to live in a large shipbuilding district, he will be able to attend
lectures preparing him for the Board of Education Higher
Stage Examination in Naval Architecture. A firstclass
certificate in this stage is worth having, and in preparing
for the examination, the student must to a large extent read
on his own account, and for that year he will be well advised
to devote his whole attention to this subject. Much will
depend on the particular arrangements of teaching adopted
in a district as to how the work can be best spread over a
series of years.
In making the above remarks, the writer wishes to empha
size the fact that a student cannot be said to learn Naval
Architecture by merely attending Naval Architecture classes.
Teachers in this subject have not the time to teach Geometry,
Applied Mechanics, or Mathematics, and unless these subjects
are familiar to the student, his education will be of a very
superficial nature. Teachers of the subject are always ready
to advise students as to the course of study likely to be most
beneficial in any given case.
Students are strongly advised to make themselves familiar
with the use of the " slide rule," which enables ship calcula
tions to be rapidly performed.
CONTENTS
CHAPTER FACE
I. AREAS, VOLUMES, WEIGHTS, DISPLACEMENT, ETC. . . I
II. MOMENTS, CENTRE OF GRAVITY, CENTRE OF BUOYANCY,
DISPLACEMENT TABLE, PLANIMETER, ETC 45
III. CONDITIONS OF EQUILIBRIUM, TRANSVERSE METACENTRE,
MOMENT OF INERTIA, TRANSVERSE BM, INCLINING
EXPERIMENT, METACENTRIC HEIGHT, ETC. ... 90
IV. LONGITUDINAL METACENTRE, LONGITUDINAL BM,
CHANGE OF TRIM 144
V. STATICAL STABILITY, CURVES OF STABILITY, CALCULA
TIONS FOR CURVES OF STABILITY, INTEGRATOR,
DYNAMICAL STABILITY 174
VI. CALCULATIONS OF WEIGHTS STRENGTH OF BUTT CON
NECTIONS, DAVITS, PILLARS, DERRICKS, SHAFT
BRACKETS 224
VII. STRAINS EXPERIENCED BY SHIPS CURVES OF LOADS,
SHEARING FORCE, AND BENDING MOMENT EQUIVA
LENT GIRDER, "SMITH" CORRECTION, TROCHOIDAL
WAVE 258
VIII. HORSEPOWER, EFFECTIVE AND INDICATED RESISTANCE
OF SHIPS COEFFICIENTS OF SPEED LAW OF COM
PARISON PROPULSION 298
IX. THE ROLLING OF SHIPS 348
X. THE TURNING OF SHIPS STRENGTH OF RUDDER HEADS 381
XI. LAUNCHING CALCULATIONS 400
APPENDIX A. SUNDRY PROOFS, TCHEBYCHEFF'S AND
BROWN'S DISPLACEMENT SHEET, ETC.,
AND MISCELLANEOUS EXAMPLES . . 406
B. TABLES OF LOGARITHMS 431
SINES, TANGENTS AND COSINES . . . 436
SQUARES AND CUBES 438
C. SYLLABUS OF NAVAL ARCHITECTURE
EXAMINATIONS 446
D. QUESTIONS AT NAVAL ARCHITECTURE
EXAMINATIONS 450
ANSWERS TO QUESTIONS 485
BIBLIOGRAPHY 488
INDEX 491
LIST OF FOLDING TABLES
At End of Book
TABLE I. DISPLACEMENT TABLE BY SIMPSON'S RULES.
PLATE I. SHEER DRAWING OF A TUG.
TABLE II. DISPLACEMENT TABLE BY JOINT RULES, TCHEBYCHEFF'S
AND SIMPSON'S.
TABLES III., IIlA. DISPLACEMENT TABLB,
TABLE IV. STABILITY TABLR.
TEXTBOOK
OF
THEORETICAL NAVAL ARCHITECTURE
CHAPTER I.
AREAS, VOLUMES, WEIGHTS, DISPLACEMENT, ETC.
Areas of Plane Figures.
A Rectangle. This is a foursided figure having its opposite
sides parallel to one another and all its angles right angles.
Such a figure is shown in D. C.
Fig. i. Its area is the pro
duct of the length and the
breadth, or AB X BC. Thus
a rectangular plate 6 feet
long and 3 feet broad will
contain
6 x 3 = 1 8 square feet FlG 
and if of such a thickness as to weigh 1 2\ Ibs. per square foot,
will weigh
18 x 12^ = 225 Ibs.
A Square. This is a particular case of the above, the
length being equal to the breadth. Thus a square hatch of
3^ feet side will have an area of
,1 * ?1 _ 7. y ?! 49.
3a * 3a 2*3 4
= 12^ square feet
Theoretical Naval Architecture.
A Triangle. This is a figure contained by three straight
lines, as ABC in Fig. 2. From the vertex C drop a perpen
dicular on to the base AB
 (or AB produced, if neces
sary). Then the area is
given by half the product
of the base into the height,
or
(AB x CD)
If we draw through the
apex C a line parallel to
the base AB, any triangle
having its apex on this line,
and navmg AB for its base, will be equal in area to the
tiiar.gle ABC. If more convenient, we can consider either A
or B as the apex, and BC or AC accordingly as the base.
Thus a triangle of base 55 feet and perpendicular drawn
from the apex z\ feet, will have for its area
= 6^ square feet
If this triangle be the form of a plate weighing 20 Ibs. to
the square foot, the weight of the plate will be
ff X20=I2 3 lbs.
A Trapezoid. This is a figure formed of four straight
lines, of which two only are
parallel. Fig. 3 gives such a
figure, ABCD.
If the lengths of the parallel
sides AB and CD are a and b
respectively, and h is the per
pendicular distance between
them, the area of the trapezoid
a, B. is g iven b y
FlG  * \(a + b) X h
or onehalf the sum of the parallel sides multiplied by the
perpendicular distance between them.
Areas, Volumes, Weights, Displacement, etc. ' 3
Example. An armour plate is of the form of a trapeoid with parallel
sides 8' 3" and 8' 9" long, and their distance apart 12 feet. Find its
weight if 6 inches thick, the material of the armour plate weighing 490 Ibs.
per cubic foot.
First we must find the area, which is given by
8' 3" + 8' 9" ^
L ^  1 X 12 square feet = # x 12
= 102 square feet
The plate being 6 inches thick = \ foot, the cubical contents of the
piate will be
102 x \  51 cubic feet
The weight will therefore be
= ii'i5 tons
A Trapezium is a quadrilateral or foursided figure of
which no two sides are parallel.
Such a figure is ABCD (Fig. 4). Its area may be found
by drawing a diagonal BD
and adding together the
areas of the triangles ABD,
BDC. These both have the
same base, BD. Therefore
from A and C drop per
pendiculars AE and CF on
to BD. Then the area of
the trapezium is given by
(AE + CF) X BD
Example. Draw a trapezium FIG 4
on scale \ inch = I foot, where
four sides taken in order are 6, 5, 6, and 10 feet respectively, and the
diagonal from the startingpoint 10 feet. Find its area in square feet.
Ans. 40 sq. feet.
A Circle. This is a figure all points of whose boundary
are equally distant from a fixed point within it called the centre.
The boundary is called its circumference^ and any line from the
centre to the circumference is called a radius. Any line passing
through the centre and with its ends on the circumference
is called a diameter.
4 Theoretical Naval Architecture.
The ratio between the circumference of a circle and its
diameter is called Tr, 1 and TT = 3*1416, or nearly ^
Thus the length of a thin wire forming the circumference
of a circle of diameter 5 feet is given by
TT x 5 = 5 X 3*1416 feet
= 15*7080 feet
or using TT = ^, the circumference = 5 x "
= ia = i 5 f feet
The circumference of a mast 2' 6" in diameter is given by
2\ X TT feet = f x ^
The area of a circle of diameter d is given by
Thus a solid pillar 4 inches in diameter has a sectional
area of
= 1 2 square inches
A hollow pillar 5 inches external diameter and inch thick
will have a sectional area obtained by subtracting the area of
a circle 4^ inches diameter from the area of a circle 5 inches
diameter
_ r ( 5 ) 2 N r (^r\
"\ 4 ) \ 4 )
= 3*73 square inches
The same result may be obtained by taking a mean
diameter of the ring, finding its circumference, and multiplying
by the breadth of the ring.
Mean diameter = 4f inches
Circumference = ~ X ^f inches
Area = (^ X  7 ) X square inches
= 3' 7 3 square inches as before
1 This is the Greek letter//, and is always used to denote 3*1416, or ? 7 2
nearly ; that is, the ratio borne by the circumference of a circle to its
diameter.
Areas, Volumes, Weights, Displacement, etc.
5
Trapezoidal Rule. 1 We have already seen (p. 2) that
the area of a trapezoid, as ABCD, Fig. 5, is given by
i(AD + BC)AB, or calling AD, BC, and AB y^ y* and h
respectively the area is given by
If, now, we have two trapezoids joined together, as in
B.
FIG. 5.
Fig. 6, having BE = AB, the area of the added part will be
given by
The area of the whole figure is given by
\(yi +y*)h + ite 4 y*)h = \^(y\ +
If we took a third trapezoid and joined on in a similar
manner, the area of the whole figure would be given by
i*c*
Trapezoidal rule for finding the area of a curvilinear figure,
as ABCD, Fig. 7.
Divide the base AB into a convenient number of equal
parts, as AE, EG, etc., each of length equal to h, say. Set up
perpendiculars to the base, as EF, GH, etc. If we join DF,
FH, etc., by straight lines, shown dotted, the area required
will very nearly equal the sum of the areas of the trapezoids
ADFE, EFHG, etc. Or using the lengths y lt y* etc., as
indicated in the figure
Area = h
+}>*
1 The Trapezoidal rule is largely used in France and in the United
States for ship calculations.
6 Theoretical Naval Architecture.
In the case of the area shown in Fig. 7, the area will be
somewhat greater than that given by this rule. If the curve,
however, bent towards the base line, the actual area would be
somewhat less than that given by this rule. In any case, the
closer the perpendiculars are taken together the less will be
the error involved by using this rule. Putting this rule into
words, we have
To find the area of a curvilinear figure, as ABCD, Fig. 7,
by means of the trapezoidal rule, divide the base into any con
venient number of equal parts, and erect perpendiculars to the base
meeting the curve ; then to the half sum of the first and last of
these add the sum of all the intermediate ones ; the result multi
plied by the common distance apart will give tJie area required.
The perpendiculars to the base AB, as AD, EF, are termed
" ordinates? and any measurement along the base from a given
startingpoint is termed an "abscissa" Thus the point P on
the curve has an ordinate OP and an abscissa AO when
referred to the point A as origin.
Simpson's First Rule. 1 This rule assumes that the
curved line DC, forming one boundary of the curvilinear area
ABCD, Fig. 8, is a portion of a curve known as a parabola of
the second order? In practice it is found that the results given
by its application to ordinary curves are very accurate, and it is
1 It is usual to call these rules Simpson's rules, but the first rule
was given before Simpson's time by James Stirling, in bis " Methodus
Differentialis," published in 1730.
2 A "parabola of the second order" is one whose equation referred
to coordinate axes is of the form^ = a, + a^x + a 2 * 2 , where a,, a lt a 2 are
constants.
Areas, Volumes, Weights, Displacement, etc.
this rule that is most extensively used in this country in finding
the areas of curvilinear figures required in ship calculations.
Let ABCD, Fig. 8, be a figure bounded on one side by
the curved line DC, which, as p
stated above, is assumed to be
a parabola of the second order.
AB is the base, and AD and
BC are end ordinates perpen
dicular to the base.
Bisect AB in E, and draw
EF perpendicular to AB, meet
ing the curve in F. Then the
area is given by
FIG. 8.
AE(AD + 4EF + BC)
or using y^y^y^ to represent the ordinates, h the common
interval between them
Now, a long curvilinear area * may be divided up into a
number of portions similar to the above, to each of which the
above rule will apply. Thus the area of the portion GHNM
of the area Fig. 7 will be given by
te
and the portion MNCB will have an area given by
"On
3
Therefore the total area will be, supposing all the ordinates
are a common distance h apart
~(y\ + 4^2 + 2^3
o
Ordinates, as GH, MN, which divide the figure into the
elementary areas are termed " dividing ordinates"
Ordinates between these, as EF, KL, OP, are termed
" intermediate ordinates"
1 The curvature is supposed continuous. If the curvature changes
abruptly at any point, this point must be at a dividing ordinate.
f
8
Theoretical Naval Architecture.
Notice that the area must have an even number of intervals^
or, what is the same thing, an odd number of ordinates^ for
Simpson's first rule to be applicable.
Therefore, putting Simpson's first rule into words, we
have
Divide the base into a convenient even number of equal parts,
and erect ordinates meeting the curve. Then to the sum of the end
ordinates add four times the even ordinates and twice the odd
ordinates. The sum thus obtained^ multiplied by onethird the
common distance apart of the ordinates^ will give the area.
Approximate Proof of Simpson's First Rule. The
truth of Simpson's first rule may be understood by the following
approximate proof : 1
Let DFC, Fig. 9, be a curved line on the base AB, and
with end ordinates AD, BC perpendicular to AB. Divide AB
equally in E, and draw the ordinate EF perpendicular to AB.
Then with the ordinary notation
Area = 
by Simpson's first rule.
Now
+ 4^
divide AB into three equal
parts by the points G and H.
Draw perpendiculars GJ and
HK to the base AB. At F
draw a tangent to the curve,
meeting GJ and HK in J and
K. Join DJ and KC. Now,
it is evident that the area we
want is very nearly equal to
the area ADJKCB. This
will be found by adding to
gether the areas of the trape
zoids ADJG, GJKH, HKCB.
+ GJ)AG
FIG. 9.
Area of ADJG =
GJKH = i(GJ + HK)GH
HKCB = i(HK + BC)HB
1 Another proof will be found on p. 77. The mathematical proof will
be found in Appendix A.
Areas, Volumes, Weights, Displacement, etc.
Now, AG = GH = HB = AB = f AE, therefore the total
area is
) (AD + 2GJ + 2HK + BC)
Now, AE = h, and GJ + HK = 2EF (this may be seen at
once by measuring with a strip of paper), therefore the total
area is
^(AD + 4EF + BC) = (y, + 4^ +.*)
J O
which is the same as that given by Simpson's first rule.
Application of Simpson's First Rule. Example. A curvi
linear area has ordinates at a common distance apart of 2 feet, the lengths
being 1*45, 2*65, 435, 645, 8*50, 1040, and 1185 feet respectively.
Find the area of the figure in square feet.
In finding the area of such a curvilinear figure by means of Simpson's
first rule, the work is arranged as follows :
Number of
ordinate.
Length of
ordinate.
Simpson's
multipliers. l
Functions of
ordinates.
2
265
4
io*6o
3
4*35
2
870
4
645
4
2580
1
850
1040
2
4
1700
4160
7
1185
I
11*85
11700 sum of functions
Common interval = 2 feet
\ common interval = feet
area =117x1 = 78 square feet
The length of this curvilinear figure is 12 feet, and it has
been divided into an even number of intervals, viz. 6, 2 feet
apart, giving an odd number of ordinates, viz. 7. We are
consequently able to apply Simpson's first rule to finding its
area. Four columns are used. In the first column are placed
the numbers of the ordinates, starting from one end of the
figure. In the second column are placed, in the proper order,
the lengths of the ordinates corresponding to the numbers in
the first column. These lengths are expressed in feet and
1 Sometimes the multipliers used are half these, viz. , 2, I, 2, I, 2, $,
and the result at the end is multiplied by twothirds the common interval.
10
Theoretical Naval Architecture.
decimals of a foot, and are best measured off with a decimal
scale. If a scale showing feet and inches is used, then the
inches should be converted into decimals of a foot ; thus,
6' g" = 675', and 6' 3^" = 6' 3'. In the next column are placed
Simpson's multipliers in their proper order and opposite their
corresponding ordinates. The order may be remembered by
combining together the multipliers for the elementary area first
considered
i 4 i
i 4 i
i 4 i
or 1424241
The last column contains the product of the length of the
ordinate and its multiplier given in the third column. These
are termed the "functions of ordinates" The sum of the
figures in the last column is termed the " sum of functions of
ordinates" This has to be multiplied by onethird the common
interval, or in this case J. The area then is given by
117 X  = 78 square feet
Simpson's Second Rule. This rule assumes that the
curved line DC, forming one boundary of the curvilinear area
H.
A E F B
FIG. 10.
ABCD, Fig. 10, is a portion of a curve known as "a parabola
of the third order" 1
Let ABCD, Fig. 10, be a figure bounded on one side by
the curved line DC, which, as stated above, is assumed to be
1 A "parabola of the third order" is one whose equation referred to
coordinate axes is of the form y = a + a^x + a^x* + a,* 8 , where a g , a^
a v fl g are constants.
Areas, Volumes, Weights, Displacement, etc. n
" a parabola of the third order" AB is the base, and AD and
BC are end ordinates perpendicular to the base. Divide the
base AB into three equal parts by points E and F, and draw
EG, FH perpendicular to AB, meeting the curve in G and H
respectively. Then the area is given by
AE(AD + 3EG + 3 FH + BC)
or, using y^ y 2 , y 3 , y^ to represent the ordinates, and h the
common interval between them
Area = \h(y^ + 372 + 3? + J^)
Now, a long curvilinear area 1 may be divided into a
number of portions similar to the above, to each of which the
above rule will apply. Thus the area of the portion KLCB in
Fig. 7 will be given by
1^4 + 3^5 + 3^6+^7)
Consequently the total area of ABCD, Fig. 7, will be,
supposing all the ordinates are a common distance h apart
The ordinate KL is termed a " dividing ordinate" and the
others, EF, GH, MN, OP, are termed " intermediate ordinates"
This rule may be approximately proved by a process similar to
that adopted on p. 8 for the first rule. 2
Application of Simpson's Second Rule. Example. A cur
vilinear area has ordinates at a common distance apart of 2 feet, the
lengths being 145, 2^65, 4*35, 6*45, 8*50, 1040, and 1185 feet respectively.
Find the area of the figure in square feet by the use of Simpson's second rule.
In finding the area of such a curvilinear figure by means of Simpson's
second rule, the work is arranged as follows :
Number of
ordinate.
Length of
ordinate.
Simpson's
multipliers.
Functions of
ordinates.
1
2
i45
265
I
3
i*45
7'95
3
4'35
3
1305
4
645
2
1 2 '9O
5
850
3
2550
6
1040
3
3120
7
1 1 '85
i
II'85
10390 sum of functions
Common interval = 2 feet
 common interval = f = f
1039 X f = 77925 square feet
1 See footnote on p. 7.
* See Appendix A for the mathematical proof!
12 Theoretical Naval Architecture.
This curvilinear area is the same as already taken for an
example of the application of Simpson's first rule. It will be
noticed that the number of intervals is 6 or a multiple of 3.
We are consequently able to apply Simpson's second rule to
finding the area. The columns are arranged as in the previous
case, the multipliers used being those for the second rule.
The order may be remembered by combining together the
multipliers for the elementary area with three intervals first
considered
or i 3 3 2 3 3 i
For nine intervals the multipliers would be i, 3, 3, 2, 3, 3,
2, 3, 3, i.
The sum of the functions of ordinates has in this case to be
multiplied by f the common interval, or f x 2 = f , and con
sequently the area is
103*9 X f = 77*925 square feet
It will be noticed how nearly the area as obtained by the
two rules agree. In practice the first rule is used in nearly all
cases, because it is much simpler than the second rule and
quite as accurate. It sometimes happens, however, that we
only have four ordinates to deal with, and in this case Simp
son's second rule must be used. When there are six ordinates,
neither of the above rules will fit. The following rule gives
the area : ff . h(\ . y, + 7a + * + y, + y* + I*). This may
be proved by applying the second rule to the middle four
ordinates, and the following 5, 8, i rule to the ends.
To find the Area of a Portion of a Curvilinear Area
contained between Two Consecutive Ordinates. Such
a portion is AEFD, Fig. 8. In order to obtain this area, we
require the three ordinates to the curve y l y?, jv 3 . The curve
DFC is assumed to be, as in Simpson's first rule, a parabola of
the second order. Using the ordinary notation, we have
Area of ADFE = ^(5^1 + 8y 2  y 3 )
Thus, if the ordinates of the curve in Fig. 8 be 8*5, 104,
Areas, Volumes, Weights, Displacement, etc. 13
1185 feet, and 2 feet apart, the area of AEFD will be given
by
y^ X 2(5 x 85 + 8 x 10*4 11*85) = 1897 square feet
Similarly the area of EBCF will be given by
^ X 2(5 x 11*85 + 8 X 104  85) = 2232 square feet
giving a total area of the whole figure as 41*29 square feet.
Obtaining this area by means of Simpson's first rule, we
should obtain 41*3 square feet. 1
This rule is sometimes known as the "fiveeight" rule.
Subdivided Intervals. When the curvature of a line
forming a boundary of an area, as Fig. n, is very sharp, it is
found that the distance apart of ordinates, as used for the
straighter part of the curve, does not give a sufficiently accurate
result. In such a case, ordinates B
are drawn at a submultiple of
the ordinary distance apart of
the main ordinates.
Take ABC, a quadrant of a
circle (Fig. n), and draw the
three ordinates y 2) y^ y^ a dis
tance h apart. Then we should
get the area approximately by
putting the ordinates through
Simpson's first rule. Now, the
curve EFC is very sharp, and
the result obtained is very far
from being an accurate one.
ordinates y\ y".
given by
Now put in the intermediate
Then the area of the portion DEC will be
or we may write this
(y 6 = o at end)
The area of the portion ABED is given by
1 See Example 25, p. 41.
14 Theoretical Naval Architecttire.
or the area of the whole figure
Thus the multipliers for ordinates onehalf the ordinary distance
apart are ^, 2, ^, and for ordinates onequarter the ordinary
distance apart are J, i, , i, \. Thus we diminish the
multiplier of each ordinate of a set of subdivided intervals in
the same proportion as the intervals are subdivided. Each
ordinate is then multiplied by its proper multiplier found in
this way, and the sum of the products multiplied by \ or f the
whole interval according as the first or second rule is used
An exercise on the use and necessity for subdivided intervals
will be found on p. 43.
Algebraic Expression for the Area of a Figure
bounded by a Plane Curve. It is often convenient to be
able to express in a short form the area of a plane curvilinear
figure.
In Fig. 12, let ABCD be a strip cut off by the ordinates
AB, CD, a distance A* apart,
A# being supposed small.
Then the area of this strip is
very nearly
y X A*
where y is the length of the
ordinate AB. If now we
imagine the strip to become
indefinitely narrow, the small
PIG j triangular piece BDE will dis
appear, and calling dx the
breadth of the strip, its area will be
y X dx
The area of the whole curvilinear figure would be found if
we added together the areas of all such strips, and this could
be written
fy.J*
where the symbol / may be regarded as indicating the sum
of all such strips as y . dx. We have already found that
Areas, Volumes, Weights, Displacement, etc. 15
Simpson's rules enable us to find the areas of such figures,
so we may look upon the expression for the area
jy.dx
as meaning that, to find the area of a figure, we take the
length of the ordinate y at convenient intervals, and put them
through Simpson's multipliers. The result, multiplied by \ or
f the common interval, as the case may be, will give the area.
A familiarity with the above will be found of great service in
dealing with moments in the next chapter.
To find the Area of a Figure bounded by a Plane
Curve and Two Radii. Let OAB, Fig. 13, be such a figure,
OA, OB being the
bounding radii.
Take two points
very close together on
the curve PP' ; join OP,
OP', and let OP = r
and the small angle
POF = A0 in circular
measure. 1 Then OP
= OP' = r very nearly,
and the area of the
elementary portion
Fio. 13.
being the length of PP',
and regarding OPP' as
a triangle. If now we consider OP, OP' to become in
definitely close together, and consequently the angle POP*
indefinitely small = dO say, any error in regarding PO P' as a
triangle will disappear, and we shall have
Area POP' =  . 46
2
and the whole area AOB is the sum of all such areas which
can be drawn between OA and OB, or
I.
1 See pp. 16 and 90.
i6
Theoretical Naval Architecture.
Now, this exactly corresponds to the algebraic expression
for the area of an ordinary plane curvilinear figure, viz.
ly.dx (seep. 15)
y corresponding to and dx corresponding to dB. Therefore
divide the angle between the bounding radii into an even
number of equal angular intervals by means of radii. Measure
these radii, and treat their halfsquares as ordinates of a curve
by Simpson's first rule, multiplying the addition by \ the
common angular interval in circular measure. Simpson's second
rule may be used in a similar manner.
The circular measure of an angle^ is the number of degrees
it contains multiplied by ^, or 001745. Thus the circular
180
measure of
9= 2
3I4I6
and the circular measure of 15 is 026175.
Example. To find the area of a figure bounded by a plane curve and
two radii 90 apart, the lengths of radii 15 apart being o, 2 '6, 5*2, 7*8, 10*5,
I3'i, 157.
Angle from
first radius.
Length of
radius.
Square of
length.
Simpson's
multipliers.
Functions of
squares.
O'O
O'O
I
O'O
I 5
26
68
4
272
30
52
270
2
54'0
45
78
608
4
2432
60
105
IIO'2
2
2204
75
131
I71'6
4
6864
900
157
2465
I
2465
14777 sum of
functions
Circular measure of 15 = 026175
/. area = 14777 X \ X 026175 X
= 64^ square feet nearly
The process is exactly the same as in Simpson's rule for a plane area
with equidistant ordinates. To save labour, the squares of the radii are put
through the proper multipliers, the multiplication by J being performed at
the end.
1 See also p. 90.
Areas, Volumes, Weights, Displacement, etc. 17
TehebyehefFs Rules. We have discussed above various
methods that can be employed for determining the area of a
figure bounded by a curved line. The methods that are most
largely employed are those known as the " Trapezoidal rule "
and " Simpson's first rule." The former is used in France and
America, 1 and the latter is used in Great Britain. The trape
zoidal rule has a great advantage in its simplicity, but con
siderable judgment is necessary in its use to obtain good
results. Simpson's first rule is rather more complex, but gives
exceedingly good results for the areas dealt with in ordinary
ship calculations.
In the above rules the spacing of ordinates is constant. A
rule has been devised for determining the area of a curvilinear
figure, in which
the multiplier is
the same for all
ordinates when
these ordinates y
are suitably /
placed so that /
the lengths of /
f
s
10
^^
o
>
&
Ol
N
\\
only need add
ing together
to obtain the
8^25 .
i
FIG.
+27*2*
ISA.
area of the figure. This rule is Tchebycheff's rule, 8 and a
much fewer number of ordinates are required than for
either the Trapezoidal or Simpson's rules for equally correct
results. For instance to find the area of a semicircle 10 feet
radius, using five ordinates. These ordinates are placed in the
positions shown in Fig. 1 3A. The lengths of these five ordinates
are found to be 554, 9*27, io'o, 9*27, 554 feet respectively, and
all that is needed is to add these lengths together, multiply the
1 See a paper read before the American Society of Naval Architects in
1895, by Mr. D. W. Taylor.
2 See a paper by Mr. C. F. Munday, M.I.N.A., in the Transactions of
the Institution of Naval Architects for 1899, and a paper by Professor Biles
in the Transactions of Institute of Engineers and Shipbuilders in Scotland
for 1899.
18
Theoretical Naval Architecture
result by the length of the figure, and divide by the number of
ordinates. The area by this rule is therefore
20
3962 X = 15848 square feet
The exact area is, of course, 157:08 square feet. Example
No. 48, Chapter I., gives an illustration of the number of
ordinates it is necessary to use for such a figure when using
Simpson's first rule in order to obtain a close approximation
to the correct area.
The following table gives the position of ordinates of a
curve with reference to the middle ordinate for different
numbers of ordinates :
No. of
ordinates
used.
Position of ordinates from middle of base in fractions of the halflength of base.
2
05773
3
o
07071
4
01876
07947
1
02666
03745
04225
08325
08662
I
O'IO26
03239
04062
05297
05938
08839
08974
9
01679
05288
06010
09116
10
00838
03127
05000
06873
09162
The following example will show how the rule is employed
to find the area of the load waterplane of a ship 600 feet long,
for which by the ordinary method 21 ordinates would have to
be used 30 feet apart. By using 10 ordinates, and drawing
them according to the above table, i.e. the following distance
from amidships both forward and aft, 25*1, 638, 150*0, 2062,
2749 feet respectively, we obtain the following lengths for the
semiordinates, commencing from forward : 3*6, 15*0, 25*1,
3i'6, 35'i, 35'4> 33'4, 28*8, 222, and 11*5 feet respectively.
These lengths are added together, and the result is multiplied
by the length of the waterplane and divided by the number of
ordinates. The result is the halfarea of the waterplane, viz.
600
2417 X = 14,502 square feet.
Areas, Volumes ', Weights, Displacement, etc. 19
The simplification due to this method consists in the fewer
number of ordinates necessary and the simple process of
addition that is required when the ordinates are measured off.
The method of proof of these rules is given in the Appendix.
Measurement of Volumes.
The Capacity or Volume of a Rectangular Block
is the product of the length, breadth, and depth, or, in other
words, the area of one face multiplied by the thickness. All
these dimensions must be expressed in the same units. Thus
the volume of an armour plate 1 2 feet long, 85 feet wide, and
1 8 inches thick, is given by
12 x 8J X 4 = 12 x ^ X f = f 1 =148! cubic feet.
The Volume of a Solid of Constant Section is the
area of its section multiplied by its length. Thus a pipe 2 feet
in diameter and 100 feet long has a section of = " square
feet, and a volume of ^ X 100 = 2M2 = 3 14^ cubic feet.
A hollow pillar 7' 6" long, 5 inches external diameter, and
\ inch thick, has a sectional area of
3*73 square inches
or  square feet
144
and the volume of material of which it is composed is
\i44/ * 2 96
= 0195 cubic foot
Volume of a Sphere. This is given by ? . d?, where d
is the diameter. Thus the volume of a ball 3 inches in dia
meter is given by
TT __ 22 x 27
6' 27= ~~^~
= i4y cubic inches
Volume of a Pyramid. This is a solid having a base
20
Theoretical Naval Architecture.
in the shape of a polygon, and a point called its vertex not in
the same plane as the base. The vertex is joined by straight
lines to all points on the boundary of the base. Its volume is
given by the product of the area of the base and onethird the
perpendicular distance of the vertex from the base. A cone is
a particular case of the pyramid having for its base a figure
with a continuous curve, and a right circular cone is a cone
having for its base a circle and its vertex immediately over the
centre of the base.
To find the Volume of a Solid bounded by a
Curved Surface. The volumes of such bodies as this are
continually required in ship calculation work, the most
important cases being the volume of the underwater portion
of a vessel. In this case, the volume is bounded on one side
by a plane surface, the waterplane of the vessel. Volumes
of compartments are frequently required, such as those for
containing fresh water or coalbunkers. The body is divided
by a series of planes
or AO.. spaced equally apart.
The area of each section
is obtained by means of
one of the rules already
explained. These areas
are treated as the ordi
nates of a new curve,
which may be run in,
with ordinates the spac
ing of the planes apart.
FIG. 14. It is often desirable to
draw this curve with
areas as ordinates as in Fig. 14, because, if the surface is a fair
surface, the curve of areas should be a fair curve, and should
run evenly through all the spots ; any inaccuracy may then be
detected. The area of the curve of areas is then obtained by
one of Simpson's rules as convenient, and this area will re
present the cubical contents of the body.
Example. A coalbunker has sections if 6" apart, and the areas of
these sections are 98, 123, 137, 135, 122 square feet respectively. Find the
Areas, Vohimes, Weights, Displacement, etc. 21
volume of the bunker and the number of tons of coal it will hold, taking
44 cubic feet of coal to weigh I ton.
Areas.
Simpson's
multipliers.
Functions of
areas.
9 8
I
9 8
I2 3
4
492
137
2
274
135
4
540
122
I
122
1526 sum of functions
 common interval = g X 175 = ^
.*. volume = 1526 x ^ cubic feet
= 8902 cubic feet,
and the bunker will hold ^ = 202 tons
The underwater portion of a ship is symmetrical about the
foreandaft middleline plane.
We may divide the volume in two ways
1. By equidistant planes perpendicular to the middleline
plane and to the load waterplane.
2. By equidistant planes parallel to the load waterplane.
The volume as obtained by both methods should be the
same, and they are used to check each other.
Examples. I. The underwater portion of a vessel is divided by vertical
sections 10 feet apart of the following areas : 03, 227, 488, 732, 884,
82*8, 587, 262, 3*9 square feet. Find the volume in cubic feet. (The
curve of sectional areas is given in Fig. 15.)
CURVE OF SECTIONAL AREAS.
\
5, 4:
FIG 15.
The number of ordinates being odd, Simpson's first rule can be applied
as indicated in the following calculation :
22
Theoretical Naval Architecture.
Number of
section.
Area of
section.
Simpson's
multipliers.
Function of
area.
I
0'3
I
0'3
2
227
4
908
3
488
2
976
4
732
4
2928
5
884
2
1768
6
828
4
331*2
7
587
2
1174
8
262
4
1048
9
3*9
1
3'9
1 2 1 5 '6 sum of functions
\ common interval = ^
/. volume = 12156 X ^
= 4052 cubic feet
2. The underwater portion of the above vessel is divided by planes
parallel to the load waterplane and I J feet apart of the following areas :
944, 795, 605, 396, 231, 120, 68, 25, 8 square feet. Find the volume in
cubic feet.
The number of areas being odd, Simpson's first rule can be applied, as
indicated in the following calculation :
Number of
waterline.
Area of
waterplane.
Simpson's
multipliers.
Function of
area.
I
944
I
944
2
795
4
3180
3
605
2
1210
4
396
4
1584
5
231
2
462
6
120
4
480
7
68
2
136
8
25
4
IOO
9
8
i
8
\ common interval =
8104 sum of functions

.*. volume = 8104 X 3
= 4052 cubic feet
which is the same result as was obtained above by taking the areas of
vertical sections and putting them through Simpson's rule.
In practice this volume is found by means of a " displace
ment sheet," or by the " planimeter." See Chapter II. and
Appendix A.
Areas, Volumes, Weights, Displacement, etc. 23
Displacement. The amount of water displaced or put
aside by a vessel afloat is termed her "displacement" This
may be reckoned as a volume, when it is expressed in cubic
feet, or as a weight, when it is expressed in tons. It is usual
to take salt water to weigh 64 Ibs. per cubic foot, and conse
quently a jJ4 Q = 35 cubic feet of salt water will weigh one
ton. Fresh water, on the other hand, is regarded as weighing
62^ Ibs. per cubic foot, or 36 cubic feet to the ton. 1 The
volume displacement is therefore 35 or 36 times the weight dis
placement, according as we are dealing with salt or fresh water.
If a vessel is floating in equilibrium in still water ^ the weight
of water she displaces miist exactly equal tht weight of the vessel
herself with everything she has on board.
That this must be true may be understood from the follow
ing illustrations :
1. Take a large basin and stand it in a dish (see Fig. 16).
Just fill the basin i ' i
to the brim with I /
water. Now care    v '  \ [~~
fully place a
smaller basin into
the water. It \~" \ x ~~7
will be found that V ^~ ~^ S
some of the water FlG  l6 
in the large basin will be displaced, and water will spill over
the edge of the large basin into the dish below. It is evident
that the water displaced by the basin is equal in amount to the
water that has been caught by the dish, and if this water be
weighed it will be found, if the experiment be conducted ac
curately, that the small basin is equal in weight to the water in
the dish that is, to the water it has displaced.
2. Consider a vessel floating in equilibrium in still water, and
imagine, if it were possible, that the water is solidified, main
taining the same level, and therefore the same density. If now
we lift the vessel out, we shall have a cavity left behind which
1 It is advisable to occasionally test the water at any particular place to
obtain the density, which may vary at different states of the tide. Thus we
have at Clydebank the water is 35*87 cubic feet to the ton ; at Dundee the
water is 1021 ozs. to cubic foot at high water, and 1006 ozs. at low water.
24 Theoretical Naval Architecture.
will be exactly of the form of the underwater portion of the
ship, as Fig. 17. Now let the cavity be filled up with water.
The amount of water we pour in will evidently be equal to the
volume of displacement of the vessel. Now suppose that the
solidified water outside again becomes liquid. The water we
have poured in will remain where it is, and will be supported
by the water surrounding it. The support given, first to the
vessel and now to the water we have poured in, by the sur
WATER SURFACE.
FIG. 17.
rounding water must be the same, since the condition of the
outside water is the same. Consequently, it follows that the
weight of the vessel must equal the weight of water poured
in to fill the cavity, or, in other words, the weight of the
vessel is equal to the weight of water displaced.
If the vessel whose displacement has been calculated on p. 22
is floating at her L.W.P. in salt water, her total weight will be
4052 435 = 115*8 tons
If she floated at the same L.W.P. in fresh water, her total weight
would be
4052 436 = 112^ tons
It will be at once seen that this property of floating bodies
is of very great assistance to us in dealing with ships. For, to
find the weight of a ship floating at a given line, we do not
need to estimate the weight of the ship, but we calculate out
from the drawings the displacement in tons up to the given line,
and this must equal the total weight of the ship.
Curve of Displacement. The calculation given on p. 22
gives the displacement of the vessel up to the loadwater plane,
but the draught of a ship continually varies owing to different
weights of cargo, coal, stores, etc., on board, and it is desirable
Areas, Volumes, Weights, Displacement, etc. 25
to have a means of determining quickly the displacement at
any given draught. From the rules we have already investi
gated, the displacement in tons can be calculated up to each
waterplane in succession. If we set down a scale of mean
draughts, and set off perpendiculars to this scale at the places
where each waterplane comes, and on these set off on a con
venient scale the displacement we have found up to that water
plane, then we should have a number of spots through which we
shall be able to pass a fair curve if the calculations are correct.
SCALE FOR DISPLACEMENT.
'OCOTONS~
FIG. 1 8,
A curve obtained in this way is termed a " curve of displacement?
and at any given mean draught we can measure the displace
ment of the vessel at that draught, and consequently know at
once the total weight of the vessel with everything she has on
board. This will not be quite accurate if the vessel is floating
at a waterplane not parallel to the designed load waterplane.
Fig. 1 8 gives a "curve of displacement" for a vessel, and the
following calculation shows in detail the method of obtaining
the information necessary to construct it.
26
TJieoretical Naval Architecture.
The areas of a vessel's waterplanes, two feet apart, are
follows :
L.W.L.
2 W.L.
3W.L.
4 W.L.
5 W.L.
6 W.L.
7 W.L.
7800 square feet.
745
6960
6290
5460
4320
2610
The mean draught to the L.W.L. is 14' o",and the displace
ment below the lowest W.L. is 7 1 tons.
To find the displacement to the L.W.L.
Number of
W.L.
Area of
waterplane.
Simpson's
multipliers.
Function of
area.
,
7800
,
7,800
2
7450
4
29,800
3
6960
2
13,920
4
6290
4
25,160
I
5460
4320
2
4
10,920
I 7 ,280
7
26lO
I
2,6lQ
107,490
\ common interval = \ X 2
/. displacement in cubic feet = 107,490 X
and displacement in tons, salt water = 107,490 X
= 2047 tons without the
appendage
Next we require the displacement up to No. 2 W.L., and
we subtract from the total the displacement of the layer between
i and 2, which is found by using the fiveeight rule as follows :
Number of
W.L.
Area of
waterplane.
Simpson's
multipliers.
Function of
area. .
I
7800
5
39,000
2
7450
8
59,600
3
6960
i
6,960
91,640
Areas, Volumes, Weights, Displacement, etc. 27
Displacement in tons between I A 2 x
No. i and No. 2 W.L.'s f = 9'> 6 4 * TW.X A
= 436 tons nearly
/. the displacement up to No. 2 )
,,, T . , > = 1611 tons without the
W.L. is 2047  436 I
appendage
The displacement between i and 3 W.L.'s can be found by
putting the areas of i, 2 and 3 W.L.'s through Simpson's first
rule, the result being 848 tons nearly.
.*. the displacement up to No.
W.L. is 2047  848
"" l S Wlth Ut
appendage
The displacement up to No. 4 W.L. can be obtained by
putting the areas of 4, 5, 6, and 7 W.L.'s through Simpson's
second rule, the result being
819 tons without the appendage
The displacement up to No. 5 W.L. can be obtained by
putting the areas of 5, 6, and 7 W.L.'s through Simpson's first
rule, the result being
482 tons without the appendage
The displacement up to No. 6 W.L. can be obtained by
means of the fiveeight rule, the result being
201 tons without the appendage
Collecting the above results together, and adding in the
appendage below No. 7 W.L., we have
Disp]
acement
up to L.W.L.
2 W.L.
4 W!L!
5 W.L.
6 W.L.
7 W.L.
2118 to
1682
1270
890
553
272
ns.
These displacements, set out at the corresponding draughts,
are shown in Fig. 18, and the fair curve drawn through forms
the "curve of displacement" of the vessel. It is usual to com
plete the curve as indicated right down to the keel, although
28
Theoretical Naval Architecture.
che ship could never float at a less draught than that given by
the weight of her structure alone, or when she was launched.
Tons per Inch Immersion. It is frequently necessary
to know how much a vessel will sink, when floating at a given
waterline, if certain known weights are placed on board, or
how much she will rise if certain known weights are removed.
Since the total displacement of the vessel must equal the weight
of the vessel herself, the extra displacement caused by putting
a weight on board must equal this weight. If A is the area
TONS PER INCH IMMERSION.
I 10 ,
FIG. 19.
of a given waterplane in square feet, then the displacement
of a layer i foot thick at this waterplane, supposing the vessel
parallelsided in its neighbourhood, is
A cubic feet .
or tons in salt water
35
For a layer i inch thick only, the displacement is
A
35 X 12
tons
Areas, Volumes, Weights, Displacement, etc. 29
and this must be the number of tons we must place on board
in order to sink the vessel i inch, or the number of tons we
must take out in order to lighten the vessel i inch. This is
termed the " tons per inch immersion " at the given waterline. 1
This assumes that the vessel is parallelsided at the waterline
for the depth of i inch up and i inch down, which may, for all
practical purposes, be taken as the case. If, then, we obtain
the tons per inch immersion at successive waterplanes parallel
to the load waterplane, we shall be able to construct a " curve
of tons per inch immersion" in the same way in which the curve
of displacement was constructed. Such a curve is shown in
Fig. 19, constructed for the same vessel for which the displace
ment curve was calculated. By setting up any mean draught,
say 1 1 feet, we can measure off the " tons per inch immersion,"
supposing the vessel is floating parallel to the load waterplane ;
in this case it is 175 tons. Suppose this ship is floating at a
mean draught of n feet, and we wish to know how much she
will lighten by burning 100 tons of coal. We find, as above,
the tons per inch to be 17^, and the decrease in draught is
therefore
ioo^i7i= 5 inches nearly
Curve of Areas of Midship Section. This curve is
sometimes plotted off on the same drawing as the displacement
curve and the curve of tons per inch immersion. The ordi
nates of the immersed part of the midship section being known,
we can calculate its area up to each of the waterplanes in
exactly the same way as the displacement has been calculated.
These areas are set out on a convenient scale at the respective
mean draughts, and a line drawn through the points thus
obtained. If the calculations are correct, this should be a fair
curve, and is known as " the curve of areas of midship section"
By means of this curve we are able to determine the area of
the midship section up to any given mean draught.
Fig. 20 gives the curve of areas of midship section for the
vessel for which we have already determined the displacement
curve and the curve of tons per inch immersion.
Coefficient of Fineness of Midship Section. If we
1 For approximate values of the "tons per inch immersion" in various
types of ships, see Example 55, p. 44.
30 Theoretical Naval Architecture.
draw a rectangle with depth equal to the draught of water at
the midship section to top of keel, and breadth equal to the
AREAS OF MID; SEC: SQI FT:
1*00. .300. ,200. 100.
FIG 20
extreme breadth at the midship section, we shall obtain what
may be termed the circumscribing rectangle of the immersed
midship section. The area of the immersed midship section
will be less than the area of this rectangle, and the ratio
area of immersed midship section
area of its circumscribing rectangle
is termed the coefficient of fineness of midship section.
Example. The midship section of a vessel is 68 feet broad at its
broadest part, and the draught of water is 26 feet. The area of the immersed
midship section is 1584 square feet. Find the coefficient of fineness of the
midship section.
Area of circumscribing rectangle = 68 X 26
= 1768 square feet
.'. coefficient = ^ = 0*895
If a vessel of similar form to the above has a breadth at
Areas t Volumes, Weights, Displacement, etc. 3 1
the midship section of 59' 6" and a draught of 22' 9", the area
of its immersed midship section will be
59^ x 22f x 0895 = I2I 3 square feet
The value of the midship section coefficient varies in
ordinary ships from about 085 to 095, the latter value being
for a section with very full section.
Coefficient of Fineness of Waterplane. This is
the ratio between the area of the waterplane and its circum
scribing rectangle.
The value of this coefficient for the load waterplane may
be taken as follows :
For ships with fine ends O'7
For ships of ordinary form 0*75
For ships with bluff ends 0*85
Block Coefficient of Fineness of Displacement.
This is the ratio of the volume of displacement to the volume
of a block having the same length between perpendiculars,
extreme breadth, and mean draught as the vessel. The
draught should be taken from the top of keel.
Thus a vessel is 380 feet long, 75 feet broad, with 27' 6"
mean draught, and 14,150 tons displacement. What is its
block coefficient of fineness or displacement ?
Volume of displacement = 14,150 X 35 cubic feet
Volume of circumscribing solid = 380 X 75 X 27^ cubic feet
.'. coefficient of fineness of 1 14150 X 35
displacement I ~~ 380 x 75 X 27^
= 0*63
This coefficient gives a very good indication of the fineness
of the underwater portion of a vessel, and can be calculated
and tabulated for vessels with known speeds. Then, if in the
early stages of a design we have the desired dimensions given,
with the speed required, we can select the coefficient of fineness
which appears most suitable for the vessel, and so determine
very quickly the displacement that can be obtained under the
conditions given.
3 2 Theoretical Naval Architecture.
Example. A vessel has to be 400 feet long, 42 feet beam, 1 7 feet draught,
and 13 \ knots speed. What would be the probable displacement ?
From available data, it would appear that a block coefficient of fineness
of 0^625 would be desirable. Consequently the displacement would be
(400 X 42 X 17 X 0*625) * 35 tons =5ioo tons about
The following may be taken as average values of the block
coefficient of fineness of displacement in various types of
ships :
Recent battleships
Recent fast cruisers
Fast mail steamers
Ordinary steamships
Cargo steamers
Sailing vessels
Steamyachts
'6O65
*5o'55
'5o*6o
'5S~'^5
'65' 80
'^5~'7S
*35*45
Prismatic Coefficient of Fineness of Displace
ment. This coefficient is often used as a criterion of the
fineness of the underwater portion of a vessel. It is the ratio
between the volume of displacement and the volume of a
prismatic solid the same length between perpendiculars as the
vessel, and having a constant crosssection equal in area to the
immersed midship section.
Example. A vessel is 300 feet long, 2100 tons displacement, and has
the area of her immersed midship section 425 square feet. What is her
prismatic coefficient of fineness ?
Volume of displacement = 2100 X 35 cubic feet
Volume of prismatic solid = 300 x 425
x
0577
Difference in Draught of Water when floating
in Sea Water and when floating in River Water.
Sea water is denser than river water ; that is to say, a given
volume of sea water say a cubic foot weighs more than the
same volume of river water. In consequence of this, a vessel,
on passing from the river to the sea, if she maintains the same
weight, will rise in the water, and have a greater freeboard
than when she started. Sea water weighs 64 Ibs. to the cubic
foot, and the water in a river such as the Thames may be
Areas, Volumes ; Weights, Displacement, etc. 33
taken, as weighing 63 Ibs. to the cubic foot. 1 In Fi^. ,3,1, let
the righthand portion represent the ship floating in river water,
and the lefthand portion represent the ship floating in salt //>#&
water. The distance between the two waterplanes will be the
amount the ship will rise on passing into sea water.
w
L
W
1
' '
W
L'
V
A. J
^ 1
B J
SEA WATER.
FIG. 21.
Let W = the weight of the ship in tons ;
T = the tons per inch immersion at the waterline
W'L' in salt water ;
/ = the difference in draught between the waterlines
WL, W'L' in inches.
Then the volume of displacement
W x 2240
in river water =  
63
in sea water
,
/. the volume of the layer =
W X 2240
~~6^~
w x 2240 w x 2*40
63
W X 2240
63 X 64
Now, the volume of the layer also = / x T x ^f^. ; there
fore we hav
22 4
W
2240 _
TT
1 See also Example 56, p. 44.
34
Theoretical Naval Architecture.
This may be put in another way. A ship, if floating in river
water, will weigh ^ less than if floating to the same waterline
in salt water. Thus, if W is the weight of the ship floating at a
given line in salt water, her weight if floating at the same line
in river water is
e^W less
and this must be the weight of the layer of displacement
between the saltwater line and the riverwater line for a given
weight W of the ship. If T be the tons per inch for salt water,
the tons per inch for river water will be ff T. Therefore the.
difference in draught will be
4 f T
W
inches, as above
Sinkage caused by a Central Compartment of a
Vessel being open to the Sea. Take the simple case of a
boxshaped vessel, ABCD, Fig. 22, floating at the waterline WL.
A. E. G.
B:
W
M
N.
L'
W. K
R
L
"
3. F 1
FIG. aa.
H.
C
This vessel has two watertight athwartship bulkheads in the
middle portion, EF and GH. A hole is made in the bottom or
side below water somewhere between these bulkheads. We
will take a definite case, and work it out in detail to illustrate
the principles involved in such a problem.
Length of boxshaped vessel
Breadth
Depth
Draught
Distance of bulkheads apart
100 feet
20
20
10
20 ,
If the vessel is assumed to be floating in salt water, its
weight must be
100 x 20 x 10
tons
Areas, Volumes, Weights, Displacement, etc. 35
Now, this weight remains the same after the bilging as
before, but the buoyancy has been diminished by the opening
of the compartment KPHF to the sea. This lost buoyancy
must be made up by the vessel sinking in the water until the
volume of displacement is the same as it originally was.
Suppose W'L' to be the new waterline, then the new volume of
displacement is given by the addition of the volumes of W'MFD
and NL'CH, or, calling d the new draught of water in feet
(40 x 20 x d)+ (40 X 20 xd) = 1600^ cubic feet
The original volume of displacement was
100 x 20 X 10 = 20,000 cubic feet
.*. 1600 d = 20,000
= 12' 6"
that is, the new draught of water is 12' 6", or the vessel will
sink a distance of 2' 6".
The problem may be looked at from another point of view.
The lost buoyancy 1320X20X10 cubic feet = 4000 cubic
feet; this has to be made up by the volumes W'MKW and
NL'LP, or the area of the intact waterplane multiplied by
the increase in draught. Calling x the increase in draught, we
shall have
80 X 20 X x = 4000
= 2' 6"
which is the same result as was obtained above.
If the bilged compartment contains stores, etc., the amount
of water which enters from the sea will be less than if the com
partment were quite empty. The volume of the lost displace
mt will then be given by the volume of the compartment up
the original waterline less the volume occupied by the
tores.
Thus, suppose the compartment bilged in the above
imple to contain coal, stowed so that 44 cubic feet of it will
reigh one ton, the weight of the solid coal being taken at
Ibs. to the cubic foot.
^ Theoretical Naval Architecture.
i cubic foot of coal, if solid, weighs 80 Ibs.
i as stowed ^ff = 51 Ibs.
Therefore in every cubic foot of the compartment there is
f cubic feet solid coal
ff space into which water will find its way
The lost buoyancy is therefore
f X 4000 =1450 cubic feet
The area of the intact waterplane will also be affected in
the same way ; the portion of the waterplane between the bulk
heads will contribute
f X 20 x 20 = 255 square feet to the area
The area of the intact waterplane is therefore
1600 1255 =185 5 square feet
The sinkage in feet is therefore
fHJ=o78, or 936 inches
In the case of a ship the same principles apply, supposing
the compartment to be a central one, and we have
Sinkage of vessel ) _ volume of lost buoyancy in cubic feet
in feet 1 ~~ area of intact waterplane in square feet
In the case of a compartment bilged which is not in the
middle of the length, change of the trim occurs. The method
of calculating this for any given case will be dealt with in
Chapter IV.
In the above example, if the transverse bulkheads EF and
GH had stopped just below the new waterline W'L', it is
evident that the water would flow over their tops, and the
vessel would sink. But if the tops were connected by a water
tight flat, the water would then be confined to the space, and
the vessel would remain afloat.
Areas, Volumes, Weights, Displacement, etc. 37
Velocity of Inflow of Water into a Vessel on
Bilging.
Let A. = area of the hole in square feet ;
d = the distance' the centre of the hole below the
surface in feet ;
v = initial rate of inflow of the water in feet per
second.
Then v %J d nearly
and consequently the volume of water
passing through the hole per second
Thus, if a hole 2 square feet in area, 4 feet below the water
line, were made in the side of a vessel, the amount of water,
approximately, that would flow into the vessel would be as
follows :
Cubic feet per second = 8 X /v/4 X 2
= 3 2
Cubic feet per minute = 32 x 60
Tons of water per minute =
35
A cub. ft.
Weights of Materials. The following table gives
average weights which may be used in calculating the weights
of materials employed in shipbuilding :
Steel 490 Ibs. per cubic foot.
Wrought iron
Cast iron ...
Copper
Brass
480
445
550
Zinc
Gunmetal
Lead
... 528
712
Elm (English)
(Canadian)
Fir (Dantzic)
Greenheart
Mahogany
(for boats)
Oak (English)
,, (Dantzic)
(African)
Pine (Pitch)
(red) ...
(yellow)
Teak ...
35
45
36
72
... 4048
35
52
47
62
40
36
30
Theoretical Naval Architecture.
It follows, from the weights per cubic foot of iron and
steel given above, that an iron plate i inch thick weighs 40 Ibs.
per square foot, and a steel plate i inch thick weighs 40*8 Ibs.
per square foot.
The weight per square foot may be obtained for other
thicknesses from these values, and we have the following :
Thickness in
Weight per square foot in
pounds.
Iron.
Steel.
\
10
102
\
15
I5'3
4
20
20 4
8
25
25'5
30
306
i
i
35
40
357
408
It is convenient to have the weight of steel per square foot
when specified in onetwentieths of an inch, as is the case in
Lloyd's rules
Thickness in
inches.
Weight per
square foot
in pounds.
Thickness in
inches.
Weight per
square foot
in pounds.
A
204
U
22'44
A
4'08
2448
i
612
1
26*52
4
816
2856
& \
1020
i$ = \
3060
I2'24
if
3264
A
1428
kl
3468
A
1632
8
3672
1=1
1836
20*40
1! = i
3876
4080
Areas, Volumes, Weights, Displacement, etc. 39
EXAMPLES TO CHAPTER I.
What is its weight if its
Ans. 95 Ibs.
*" IG  2 3
1. A plate has the form shown in Fig. 23.
weight per square foot is 10 Ibs. ?
2. The material of an
armour plate weighs 490 Ibs.
a cubic foot. A certain
plate is ordered 400 Ibs. per
square foot : what is its
thickness ?
Ans. 9 '8 inches.
3. Steel armour plates,
as in the previous question,
are ordered 400 Ibs. per
square foot instead of 10
inches thick. What is the
saving of weight per 100
square feet of surface of this
armour ?
Ans. 833 Ibs., or 037 ton.
4. An iron plate is of the dimensions shown in Fig. 24. What is its area ?
If two lightening holes 2' 3" in diameter are cut in it, what will its
area then be ? .
Ans. 33! square feet ; i
258 square feet. . , j
5. A hollow pillar is 4 inches L _____ ^
external diameter and jj inch
thick. What is its sectional
area, and what would be the
weight in pounds of 10 feet of
this pillar if made of wrought
iron ?
Ans. 427 square inches ;
142 Ibs.
6. A steel plate is of the
form and dimensions shown in Fig. 25. What is its weight ? (A steel plate
\ inch thick weighs 25*5 Ibs. per square foot.)
Ans. 1267 Ibs.
\
A 4
\
M
rd
FlG .
FIG. 25.
7. A wroughtiron armour plate is 15' 3'
thick. Calculate its weight in tons.
long, 3' X 6" wide, and4i inches
Ans. 429 tons.
4 Theoretical Naval Architecture.
8. A solid pillar of iron of circular section is 6' 10" long and 2\ inches
in diameter. What is its weight ?
Ans. 90^ Ibs.
9. A Dantzic fir deck plank is 22 feet long and 4 inches thick, and
tapers in width from 9 inches at one end to 6 inches at the other. What is
its weight ?
Ans. 165 Ibs.
10. A solid pillar of iron is 7' 3" long and 2f inches diameter. What
is its weight ?
Ans. 143 Ibs.
11. The total area of the deck plan of a vessel is 4500 square feet.
What would be the surface of deck plank to be worked, if there are
4 hatchways, each 4' X 2$'
2 ,, ,, 10' X 6'
and two circular skylights, each 4 feet in diameter, over which no plank is
to be laid ?
Ans. 431486 square feet.
12. A pipe is 6 inches diameter inside. How many cubic feet of water
will a length of 100 feet of this pipe contain?
Ans. 1 9 '6 cubic feet.
13. A mast 90 feet in length and 3 feet external diameter, is composed of
20 Ib. plating worked flushjointed on three Tbars, each 5" x 3" X iS^lbs.
per foot. Estimate the weight, omitting straps, and rivet heads.
Ans. 9^ tons nearly.
14. A curve has the following ordinates, 1*4" apart: IO'86, 13*53, I4'58,
15*05, 15*24, 15*28, 15*22 feet respectively. Draw this curve, and find
its area
(1) By Simpson's first rule ;
(2) By Simpson's second rule.
Ans. (i) 11607 square feet; (2) 116*03 square feet.
15. The semiordinates in feet of a vessel's midship section, starting
from the load waterline, are 26*6, 26*8, 26*8, 26*4, 254, 234, and 18*5 feet
respectively, the ordinates being 3 feet apart. Below the lowest ordinate
there is an area for one side of the section of 24*6 square feet. Find the
area of the midship section, using
(1) Simpson's first rule ;
(2) Simpson's second rule.
Ans. (i) 961 square feet ; (2) 9607 square feet.
16. The internal dimensions of a tank for holding fresh water are
8' o" X 3' 6" X 2' 6". How many tons of water will it contain ?
Ans. I '94.
17. The yfo^ordinates of a deck plan in feet are respectively ij, 5^,
10}, 13^, 14!, 14!, I2j, 9, and 3^, and the length of the plan is 128 feet.
Find the area of the deck plan in square yards.
Ans. 296.
1 8. Referring to the previous question, find the area in square feet of the
portion of the plan between the ordinates ij and 5^.
Ans. 106*7.
19. The halfordinates of the midship section of a vessel are 22*3, 22*2,
217, 20'6, 17*2, 13*2, and 8 feet in length respectively. The common
interval between consecutive ordinates is 3 feet between the first and fifth
ordinates, and i' 6" between the fifth and seventh. Calculate the total area
of the section in square feet.
Ans. 586'2 square feet.
Areas, Volumes, Weights, Displacement, etc. 4 1
20. Obtain the total area included between the first and fourth ordinates
of the section given in the preceding question.
Ans. 392 '8 square feet.
21. The semiordinates of the load waterplane of a vessel are 0*2, 36,
74, 10, n, 107, 93, 6'5, and 2 feet respectively, and they are 15 feet
apart. What is the area of the load water plane ?
Ans. 1808 square feet.
22. Referring to the previous question, what weight must be taken out
of the vessel to lighten her 3 inches ?
What additional immersion would result by placing 5 tons on board ?
Ans. 15 tons ; I'i6 inch.
23. The " tons per inch immersion " of a vessel when floating in salt
water at a certain waterplane is 44' 5. What is the area of this plane ?
Ans. 18,690 square feet.
24. A curvilinear area has ordinates 3 feet apart of length 97, 10*0, and
13*3 feet respectively. Find
(1) The area between the first and second ordinates.
(2) The area between the second and third ordinates.
(3) Check the addition of these results by finding the area of the whole
figure by Simpson's first rule.
25. Assuming the truth of the fiveeight rule for finding the area between
two consecutive ordinates of a curve, prove the truth of the rule known as
Simpson's first rule.
26. A curvilinear area has the following ordinates at equidistant intervals
of 18 feet : 6'2O, 13*80, 2190, 26*40, 22*35, I 4'7 O > and 735 feet.
Assuming that Simpson's first rule is correct, find the percentage of error
that would be involved by using
(1) The trapezoidal rule ;
(2) Simpson's second rule.
Ans. (i) i'2 per cent. ; (2) 0*4 per cent.
27. A compartment for containing fresh water has a mean section of
the form shown in Fig. 26. The length 9
of the compartment is 12 feet. How many 88.
tons of water will it contain ?
Ans. 17 tons.
28. A compartment 20 feet long, 20
feet broad, and 8J feet deep, has to be
lined with teak 3 inches in thickness.
Estimate the amount of teak required in
cubic feet, and in tons.
Ans. 365 cubic feet; 8"6 tons.
29. The areas of the waterline sec
tions of a vessel in square feet are re
spectively 2000, 2000, 1600, 1250, and
300. The common interval between them
is ij foot. Find the displacement of the
vessel in tons in salt water, neglecting the
small portion below the lowest waterline FIG. a6.
section.
Ans. 264! tons.
30. A series of areas, 17' 6" apart, contain 0^94, 2*08, 3*74, 5'33, 827,
1214, 1696, 2182, 2468, 2466, 2256, 1790, 1266, 840, 569, 373,
2 "6 1, 2 "06, o square feet respectively. Find the volume of which the above
are the sectional areas.
Ans. 3429 cubic feet.
4 2 Theoretical Naval Architecture.
31. Show how to estimate the change in the mean draught of a vessel in
going from salt to river water, and vice versd.
A vessel floats at a certain draught in river water, and when floating in
sea water without any change in lading, it is found that an addition of 175
tons is required to bring the vessel to the same draught as in river water
What is the displacement after the addition of the weight named ?
Ans. 11,200 tons.
32. The vertical sections of a vessel 10 feet apart have the following
areas : 10, 50, 60, 70, 50, 40, 20 square feet. Find the volume of displace
ment, and the displacement in tons in salt and fresh water.
Ans. 2966 cubic feet ; 84*7 tons, 82*4 tons.
33. A cylinder is 500 feet long, 20 feet diameter, and floats with, the
axis in the waterline. Find its weight when floating thus in salt water.
What weight should be taken out in order that the cylinder should float
with its axis in the surface if placed into fresh water ?
Ans. 2244 tons ; 62 tons.
34. A vessel is 500 feet long, 60 feet broad, and floats at a mean draught
of 25 feet when in salt water. Make an approximation to her draught
when she passes into river water. (Coefficient of displacement. O'5 : coefficient
ofL.W.P., 0'6.)
Ans. 25' 4".
35. A piece of teak is 20 feet long, 4$ inches thick, and its breadth
tapers from 12 inches at one end to 9 inches at the other end. What is its
weight, and how many cubic feet of water would it displace if placed into
fresh water (36 cubic feet to the ton) ?
Ans. 348 Ibs. ; 5$ cubic feet about.
36. The area of a waterplane is 5443 square feet. Find the tons per
inch immersion. Supposing 40 tons placed on board, how much would the
vessel sink ?
State any slight error that may be involved in any assumption made. If
40 tons were taken out, would the vessel rise the same amount ? What
further information would you require to give a more accurate answer ?
Ans. 12*96 tons; 3*1 inches nearly.
37. Bilge keels are to be fitted to a ship whose tons per inch is 48.
The estimated weight of the bilge keels is 36 tons, and the volume they
occupy is 840 cubic feet. What will be the increase of draught due to
fitting these bilge keels?
Ans. \ inch.
38. The tons per inch of a vessel at waterlines 2 feet apart are 1945,
1851, 1725, 156, 1355, 1087, and 652, the lowest waterline being 18
inches above the underside of flat keel. Draw the curve of tons per inch
immersion to scale, and estimate the number of tons necessary to sink the
vessel from a draught of 12 feet to a draught of 13' 6".
Ans. 344 tons.
39. The steamship Umbria is 500 feet long, 57 feet broad, 22' 6"
draught, 9860 tons displacement, 1 150 square feet area of immersed midship
section. Find
(1) Block coefficient of displacement.
(2) Prismatic ,, ,,
(3) Midshipsection coefficient.
Ans. (i) 0538; (2) 06 ; (3) 0896.
40. The steamship Orient is 445 feet long, 46 feet broad, 21' 4^" draught
mean ; the midship section coefficient is 0919, the block coefficient of dis
placement is 0621. Find
Areas, Volumes, Weights, Displacement, etc. 43
(1) Displacement in tons.
(2) Area of immersed midship section.
(3) Prismatic coefficient of displacement.
Ans. (i) 7763 tons ; (2) 904 square feet ; (3) 0675.
41. A steam yacht is 144 feet long, 22' 6" broad, 9 feet draught ; dis
placement, 334 tons salt water j area of midship section, 124 square ieet.
Find
(1) Block coefficient of displacement.
(2) Prismatic
(3) Midshipsection coefficient.
Ans. (i) 04; (2) 0655; (3) 0612.
42. Find the displacement in tons in salt water, area of the immersed
midship section, prismatic coefficient of displacement, having given the
following particulars : Length, i68feet ; breadth, 25 feet ; draught, lo'6";
midshipsection coefficient, 0*87 ; block coefficient of displacement, 0505.
Ans. 750 tons ; 2285 square feet ; 0*685.
43. A vessel in the form of a box, 100 feet long, 10 feet broad, and 20
feet deep, floats at a draught of 5 feet. Find the draught if a central
compartment IO feet long is bilged below water.
Ans. 5' 6J".
44. In a given ship, pillars in the hold can be either solid iron 4! inches
diameter, or hollow iron 6 inches diameter and half inch thick. Find the
saving in weight for every 100 feet length of these pillars, if hollow pillars
are adopted instead of solid, neglecting the effect of the solid heads and
heels of the hollow pillars.
Ans. i 35 ton.
45. What is the solid contents of a tree whose girth (circumference) is
60 inches, and length is 18 feet?
Ans. 3 5 '8 cubic feet nearly.
46. A portion of a cylindrical steel stern shaft casing is I2f feet long,
ii inch thick, and its external diameter is 14 inches. Find its weight in
pounds.
Ans. 2170 Ibs.
47. A floating body has a waterplane whose semiordinates 25 feet
apart are 03, 8, 12, 10, 2 feet respectively, and every square station is in
the form of a circle with its centre in the waterplane. Find the volume of
displacement (TT = 3/&).
Ans. 12,414 cubic feet.
48. A quadrant of 16 feet radius is divided by means of ordinates parallel
to one radius, and the following distances away : 4, 8, 10, 12, 13, 14, 15
feet respectively. The lengths of these ordinates are found to be 1549,
1386, 1249, 1058, 933, 775, and 557 feet respectively. Find
(1) The exact area to two places of decimals.
(2) The area by using only ordinates 4 feet apart.
(3) The area by using also the halfordinates.
(4) The area by using all the ordinates given above.
(5) The area as accurately as it is possible, supposing the ordinate 1249
had not been given.
Ans. (i) 20106; (2) 197*33; (3) 19975 5 (4) 20059; (5) 20050.
49. A cylindrical vessel 50 feet long and 16 feet diameter floats at a
constant draught of 12 feet in salt water. Using the information given in
the previous question, find the displacement in tons.
Ans. 231 tons nearly.
50. A bunker 24 feet long has a mean section of the form of a trapezoid,
with length of parallel sides 3 feet and 4*8 feet, and distance between them
10'5 feet. Find the number of tons of coal contained in the bunker, assuming
44 Theoretical Naval Architecture.
I ton to occupy 43 cubic feet. If the parallel sides are perpendicular to
one of the other sides, and the side 4*8 feet long is at the top of the section,
where will the top of 17 tons of coal be, supposing it to be evenly
distributed ?
(This latter part should be done by a process of trial and error.)
Ans. 228 tons ; 2' 3" below the top.
51. The sections of a ship are 20 feet apart. A coalbunker extends
from 9 feet abaft No. 8 section to I foot abaft No. 15 section, the total
length of the bunker thus being 132 feet. The areas of sections of the
bunker at Nos. 8, II, and 15 are found to be 126, 177, and 145 square
feet respectively. With this information given, estimate the capacity of the
bunker, assuming 44 cubic feet of coal to go to the ton. Stations numbered
from forward.
Ans. 495 tons.
52. The tons per inch immersion at waterlines 2 feet apart are 1809,
i6'8o, 15*15, I3'I5, I0 4 49, and 648. The draught of water to the top
waterline is n' 6", and below the lowest waterline there is a displacement
of 75 '3 tons. Find the displacement in tons, and construct a curve of
displacement.
Ans. 1712 tons.
53. A tube 35 feet long, 16 feet diameter, closed at the ends, floats in
salt water with its axis in the surface. Find approximately the thickness of
the tube, supposed to be of iron, neglecting the weight of the ends.
Ans. 0*27 foot.
54. Find the floating power of a topmast, length 64 feet, mean diameter
21 inches, the wood of the topmast weighing 36 Ibs. per cubic foot.
(The floating power of a spar is the weight it will sustain, and this is
the difference between its own weight and that of the water it displaces.
In constructing a raft, it has to be borne in mind that all the weight of
human beings is to be placed on it, and that a great quantity of provisions
and water may be safely carried Hinder it. For instance, a cask of beef
slung beneath would be 116 Ibs., above 300 Ibs. See " Sailor's Pocket
book," by Admiral Bedford.)
Ans. 43 10 Ibs.
55. Show that the following approximate values may be taken for the
" tons per inch immersion " in salt water at the load draught :
(1) For ships with fine ends ^ X L X B
(2) ,, of ordinary form 555 X L X B
(3) with bluff ends 5 fo X L X B
L and B being the length and breadth respectively of the load waterplane.
56. Show that a vessel passing from water of density d' into water^ of
density d (<f being greater than d) will decrease her freeboard by 7p j
inches, where W is the displacement in tons and T the tons per inch
immersion when in the denser water.
A vessel 400 feet long, 45 feet broad, floats in Belfast water (ion ozs.
to a cubic foot) at a draught of 21' 2^". By how much will the free
board be increased when in salt water (1025 ozs. to a cubic foot) ? (Coeffi
cient of fineness of displacement, 0*62 ; coefficient of fineness of L.W.P.,.
07S)
Ans. 2 '9 inches.
CHAPTER IL
MOMENTS, CENTRE OF GRAVITY, CENTRE OF BUOY
ANCY, DISPLACEMENT TABLE, PLANIMETER, ETC.
Principle of Moments. The moment of a force about
any given line is the product of the force into the perpen
dicular distance of its line of action from that line. It may
also be regarded as the tendency to turn about the line. A
man pushes at the end of a capstan bar (as Fig. 27) with a
w.
FIG. 7
certain force. The tendency of the capstan to turn about its
axis is given by the force exerted by the man multiplied by
his distance from the centre of the capstan, and this is the
moment of the force about the axis. If P is the force exerted
by the man in pounds (see Fig. 27), and d is his distance from
the axis in feet, then
The moment about the axis = P x d footlbs.
The same moment can be obtained by a smaller force with
a larger leverage, or a larger force with a smaller leverage, and
the moment can be increased :
(1) By increasing the force;
(2) By increasing the distance of the force from the axis.
4 6
Theoretical Naval Architecture,
If, in addition, there is another man helping the first man,
exerting a force of F Ibs. at a distance from the axis of <?
feet, the total moment about the axis is
We must now distinguish between moments tending to turn
one way and those tending to turn in the opposite direction.
Thus, in the above case, we may take a rope being wound
on to the drum of the capstan, hauling a weight W Ibs. If the
radius of the drum be a feet, then the rope tends to turn the
capstan in the opposite direction to the men, and the moment
about the axis is given by
W X a footlbs.
If the weight is just balanced, then there is no tendency to
turn, and hence no moment about the axis of the capstan, and
leaving out of account all consideration of friction, we have
(P X <*) + (F X O = W x a
The most common forces we have to deal with are those
caused by gravity, or the attraction of bodies to the earth. This
is known as their weight, and the direction of these forces must
gft be parallel at any given place. If we have a number of
weights, Wi, W a , and W 8 , on a beam at A, B, and C (Fig. 28),
n
W.
FIG. 28.
whose end is fixed at O, the moment of these weights about O
is given by
(Wj X AO) + (W. X BO) + (W, X CO)
This gives the tendency of the beam to turn about O, due to
Moments, Centre of Gravity, Centre of Buoyancy, etc. 47
the weights W t , W 2 , and W 3 placed upon it, and the beam must
be strong enough at O in order to resist this tendency, or, as
it is termed, the bending moment. Now, we can evidently
place a single weight W, equal to the sum of the weights
W 1} W 2 , and W 3 , at some point on the beam so that its moment
about O shall be the same as that due to the three weights.
If P be this point, then we must have
W x OP = (W t X OA) + (W 2 x OB) + (W 8 x OC)
or, since W = W a + W 3 + W,
OP = (W t X OA) + (W 2 x OB) + (W 8 x OC)
Wl + W 2 + W,
Example. Four weights, 30, 40, 50, 60 Ibs. respectively, are placed
on a beam fixed at one end, O, at distances from O of 3, 4, 5, 6 feet
respectively. Find the bending moment at O, and also the position of a
single weight equal to the four weights which will give the same bending
moment.
Bending moment at O = (30 X 3) + (40 X 4) + (50 X 5) + (60 x 6)
= 90 + 160 + 250 + 360
= 860 footlbs.
Total weight = 180 Ibs.
.'. position of single weight = f($ = 4$ feet from O
Centre of Gravity. The single weight W above, when
placed at P, has the same effect on the beam at O as the
three weights W x , W a , and W 8 . The point P is termed the
centre of gravity of the weights W 1} W 2 , and W s . Thus we
may define the centre of gravity of a number of weights as
follows :
The centre of gravity of a system of weights is that point at
which we may regard the whole system as being concentrated.
This definition will apply to the case of a solid body, since
we may regard it as composed of a very large number of small
particles, each of which has a definite weight and occupies
a definite position. A homogeneous solid has the same
density throughout its volume ; and all the solids with which
we have to deal are taken as homogeneous unless otherwise
specified.
It follows, from the above definition of the centre of
gravity, that if a body is suspended at its centre of gravity,
48 Theoretical Naval Architecture.
it would be perfectly balanced and have no tendency to move
away from any position in which it might be placed.
To Find the Position of the Centre of Gravity
of a number of Weights lying in a Plane. Two lines
are drawn in the plane at right angles, and the moment of the
system of weights is found successively about each of these
lines. The total weight being known, the distance of the
centre of gravity from each of these lines is found, and conse
quently the position of the centre of gravity definitely fixed.
FIG.
The following example will illustrate the principles in
volved : Four weights, of 15, 3, 10, and 5 Ibs. respectively,
are lying on a table in definite positions as shown in Fig. 29.
Find the position of the centre of gravity of these weights.
(If the legs of the table were removed, this would be the place
where we should attach a rope to the table in order that it
should remain horizontal, the weight of the table being
neglected.)
Moments, Centre of Gravity p , Centre of Buoyancy, etc. 49
Draw two lines, Ox, Oy, at right angles on the table in
any convenient position, and measure the distances of each of
the weights from Ox, Oy respectively : these distances are
indicated in the figure. The total weight is 33 Ibs. The
moment of the weights about Ox is
(i5 X 7) + (3 X 3) + (10 X 5) + (5 X 15) = 1715 footlbs.
The distance of the centre of gravity from Ox = = 5*2 feet
If we draw a line A A a distance of 5*2 feet from Ox, the
centre of gravity of the weights must be somewhere in the
line AA.
Similarly, we take moments about Oy, finding that the
moment is 150 footlbs., and the distance of the centre of
gravity from Oy is
V# = 4*25 ^et
If we draw a line BB a distance of 4*25 feet from Oy, the
centre of gravity of the weights must be somewhere in the line
BB. The point G, where AA and BB meet, will be the centre
of gravity of the weights.
Centres of Gravity of Plane Areas. A plane area has
length and breadth, but no thickness, and in order to give a
definite meaning to what is termed its centre of gravity, the
area is supposed to be the surface of a thin lamina or plate of
homogeneous material of uniform thickness. With this sup
position, the centre of gravity of a plane area is that point at
which it can be suspended and remain in equilibrium.
Centres of Gravity of Plane Figures.
Circle. The centre of gravity of a circle is obviously at
its centre.
Square and Rectangle. The centre of gravity of
either of these figures is at the point where the diagonals
intersect.
Rhombus and Rhomboid. The centre of gravity of
either of these figures is at the point where the diagonals
intersect.
Theoretical Naval Architecture.
D.
FIG. 30.
Triangle. Take the triangle ABC, Fig. 30. Bisect any
two sides BC, AC in the points D and E. Join AD, BE. The
point G where these two lines intersect is the centre of gravity
of the triangle. It can be
proved that the point G is
situated so that DG is onethird
DA, and EG is onethird EB.
We therefore have the following
rules :
i. Bisect any two sides of
the triangle, and join the points
thus obtained to the opposite angu
lar points. Then the point in
which these two lines intersect is the centre of gravity of the triangle.
2. Bisect any side of the triangle, and join the point thus
obtained with the opposite angular point. The centre of gravity
of the triangle will be on this line, and at a point at onethird its
length measured from the bisected side.
Trapezium. Let ABCD, Fig. 31, be a trapezium. By
joining the corners A and C we can divide the figure into two
triangles, ADC, ABC. The centres of gravity, E and F, of
these triangles can be
found as indicated
above. Join EF. The
centre of gravity of the
whole figure must be
somewhere in the line
EF. Again, join the
corners D and B, thus
dividing the figure into
two triangles ADB,
CDB. The centres of
triangles can be found. The
D. C
FIG. 31.
gravity, H and K, of these
centre of gravity of the whole figure must be somewhere in the
line HK ; therefore the point G, where the lines HK and EF
intersect, must be the centre of gravity of the trapezium.
The following is a more convenient method of finding the
centre of gravity of a trapezium.
Moments, Centre of Gravity, Centre of Buoyancy, etc. 5 l
Let ABCD, Fig. 32, be a trapezium. Draw the diagonals
AC, BD, intersecting at E. In the figure CE is greater than
a
FIG. 39.
EA, and DE is greater than EB. Make CH = EA and DF
= EB. Join FH. Then the centre of gravity of the triangle
EFH will also be the centre of gravity of the trapezium ABCD. 1
(A useful exercise in drawing would be to take a trapezium
on a large scale and find its
centre of gravity by each of
the above methods. If the
drawing is accurately done, the
point should be in precisely
the same position as found by
each method.)
To find the Centre of
Gravity of a Plane Area
by Experiment. Draw out
the area on a piece of card
board or stiff paper, and cut
out the shape. Then suspend
the cardboard as indicated in
Fig 33, a small weight, W,
being allowed to hang plumb.
A line drawn behind the string AW must pass through the
centre of gravity. Mark on the cardboard two points on the
string, as A and B, and join. Then the centre of gravity must
lie on AB. Now suspend the cardboard by another point, C,
1 See Example 22, for C.G. of a trapezoid.
Theoretical Naval Architecture.
as in Fig. 34, and draw the line CD immediately behind the
string of the plumbbob W. Then also the centre of gravity
must lie on the line CD. Consequently it follows that the
point of intersection G of the
lines AB and CD must be the
centre of gravity of the given
area.
Example. Set out the section of
a beam on a piece of stiff paper, and
find by experiment the position of its
centre of gravity, the beam being formed
of a bulb plate 9 inches deep and
i inch thick, having two angles on the
upper edge, each 3" x 3" X ".
Ans. 3 inches from the top.
Centres of Gravity of Solids
formed of Homogeneous
Material.
Sphere. The centre of
gravity of a sphere is at its centre.
Cylinder. The centre of
gravity of a cylinder is at one
half its height from the base, on
the line joining the centres of
gravity of the ends.
Pyramid or Cone. The centre of gravity of a pyramid
or cone is at one fourth the height of the apex from the base,
on the line joining the centre of gravity of the base to the
apex.
Moment of an Area.
The geometrical moment of a plane area relatively to a
given axis, is the product of its area into the perpendicular
distance of its centre of gravity from the given axis. It follows
that the position of the centre of gravity is known relatively to
the given axis if we know the geometrical moment about the
axis and also the area, for the distance will be the moment
divided by the area. It is usnal to speak of the moment of an
Moments, Centre of Gravity, Centre of Buoyancy, etc. 53
area about a given axis when the geometrical moment is really
meant.
To find the Position of the Centre of Gravity of
a Curvilinear Area with respect to one of its Ordi
nates. Let AEDO, Fig. 35, be a plane curvilinear area, and
we wish to find its centre of gravity with respect to the end
ordinate, OA. To do this, we must first find the moment of
the total area about OA, and this divided by the area of the
figure itself will give the distance of the centre of gravity from
OA. Take any ordinate, PQ, a distance of x from OA, and
at PQ draw a strip A* wide. Then the area of the strip is
y x A# very nearly, and the moment of the strip about OA is
(y x &x)x very nearly.
If now A* be made indefinitely small, the moment of the
strip about OA will be
y . x . dx
Now, we can imagine the whole area divided up into such
strips, and if we added up the moments about OA of all such
strips, we should obtain the total moment about OA. Therefore,
using the notation we employed for finding the area of a plane
curvilinear figure on p. 14, we shall have
Moment of the total area about OA = jy . x . dx
The expression for the area i
54
Theoretical Naval Architecture.
and this is of the same form as the expression for the moment.
Therefore, instead of y we put yx through Simpson's rule in
the ordinary way, and the result will be the moment about OA.
Set off on BC a length BF = BC X h, and on DE a length
DG = DE x 2h. Then draw through all such points a curve,
as OFG. 1 Any ordinate of this curve will give the ordinate of
the original curve at that point multiplied by its distance from
OA. The area of an elementary strip of this new curve will be
y . x . dx, and the total area of the new curve will be jy . x . dx,
or the moment of the original figure about OA. Therefore, to
find the moment of a curvilinear figure about an end ordinate,
we take each ordinate and multiply it by its distance from the
end ordinate. These products, put through Simpson's rule in
the ordinary way, will give the moment of the figure about the
end ordinate. This moment divided by the area will give the
distance of the centre of gravity of the area from the end
ordinate.
Example. A midship section has semiordinates, i' 6" apart, com
mencing at the L.W.L., of length 8'6o, 8'io, 695, 490, 275, 150, 070
feet respectively. Find the area of the section and the distance of its C.G.
from the L.W.L,
Number of
ordinates.
Length of
ordinates.
Simpson's
multipliers.
Function of
ordinates.
Number of
intervals from
No. i ord.
Products for
moment.
,
8'60
,
8'60
O'O
2
810
4
3240
I
3240
3
6'95
2
I300
2
2780
4
490
4
1960
3
5880
j
275
2
S'SO
4
2200
6
150
4
6"oo
5
3000
7
070
1
070
6
420
8670
Function of
moment
;2o
The halfarea will be given by 8670 X (3 X 1*5) = 43'35 square feet
and the whole area is 86*70
The arrangement above is adopted in order to save labour in
1 If the original curve is a parabola of the second order, this curve will
be one of the third order, and it can be proved that Simpson's first rule
will integrate a parabola of the third order.
Moments ', Centre of Gravity, Centre of Buoyancy, etc. 55
finding the moment of the area. In the fourth column we have
the functions of the ordinates, or the ordinates multiplied succes
sively by their proper multipliers. In the fifth column is placed,
not the actual distance of each ordinate from the No. i ordi
nate, but the number of intervals away, and the distance apart
is brought in at the end. In the sixth column the products of
the functions in column 4 and the multipliers in column 5 are
placed. It will be noticed that we have put the ordinates
through Simpson's multipliers first, and then multiplied by the
numbers in the fifth column after. This is the reverse to the
rule given in words above, which was put into that form in
order to bring out the principle involved more plainly. The
final result will, of course, be the same in either case, the
method adopted giving the result with the least amount of
labour, because column 4 is wanted for finding the area. The
sum of the products in column 6 will not be the moment
required, because it has to be multiplied as follows : First, by
onethird the common interval, and second, by the distance
apart of the ordinates.
The moment of the halfarea )
about the L.W.L. I = W 2 X X '*> X *
and the distance of the C.G. of the halfarea from the L.W.L.
is
1 31*4
Moment r area =*   = 303 feet
43'35
It will be noticed that we have multiplied both columns
4 and 6 by onethird the common interval, the distance of the
C.G. from No. i ordinate being obtained by
17520 X (JX 15) X 15
8670 X ft X 15)
The expression ^ X 1*5 is common to both top and bottom,
and so can be cancelled out, and we have
17520 x i'5
^7 * = 303 feet
56 Theoretical Naval Architecture.
The position of the centre of gravity of the halfarea with
regard to the L.W.L. is evidently the same as that of the whole
area.
When finding the centre of gravity of a large area, such as
a waterplane of a vessel, it is usual to take moments about
the middle ordinate. This considerably simplifies the work,
because the multipliers in column 5 are not so large.
Example. T^e semiordinates of the load waterplane of a vessel 395
feet long are, commencing from forward, o, 10*2, 20*0, 27*4, 32*1, 34*0",
33'8, 317, 276, 206, 94. Find the area and the distance of its C.G.
from the middle ordinate.
In addition to the above, there is an appendage abaft the last ordinate,
having an area of 153 square feet, and whose C.G. is 5 '6 feet abaft the last
ordinate. Taking this appendage into account, find the area and the
position of the C.G. of the waterplane.
Number of
ordinates.
Length of
ordinates.
Simpson's
multipliers.
Function of
ordinal.es.
Number of
interval from
mid. ord.
Product for
moment.
,
O'O
I
0'0
5
O'O
2
IO"2
4
408
4
1632
3
20'0
2
40  o
3
I2O'O
4
27 '4
4
io9'6
2
2I92
S
3 2 I
2
642
I
6 4 2
6
34'0
4
1360
O
5666
7
338
2
676
I
676
8
317
4
1268
2
253*6
9
276
2
55 '2
3
165 6
10
20 '6
4
824
4
3296
ii
9'4
I
9'4
5
470
7320 8634
The halfarea will be given by
7320 X (J X 395; = 9638 square feet
The fifth column gives the number of intervals away from the middle
ordinate, and the products are obtained for the forward portion adding up
to 566*6, and they are obtained for the after portion adding up to 863*4.
This gives an excess aft of 8634 5666 = 296*8. The distance of the C.G,
abaft the middle ordinate is then given by
The area of both sides is 19,276 square feet.
The second part of the question takes into account an appendage abaft
No. 1 1 ordinate, having an area of 153 square feet.
Moments, Centre of Gravity, Centre of Buoyancy, etc. 57
The total area will then be
19,276 + 153 = 19,429 square feet
To find the position of the C G. of the whole waterplane, we take
moments about ND. 6 ordinate, the distance of the C.G. of the appendage
from it being
1 97'S + 5' 6 = 2 3'i feet
Moment of main area abaft No. ordinate = 19,276 X i6'oi = 308,609
appendage = 153 X 2031 = 31,074
/.total moment abaft No. 6 ordinate = 308,609 f 31,074
and the distance of the centre of gravity \ 339683 _ .
of the whole area abaft No. 6 ordinate / = I9429 ~ r 7'4 i
To find the Position of the Centre of Gravity of
a Curvilinear Area contained between Two Con
secutive Ordinates with respect to the Near End
Ordinate. The rule investigated in the previous paragraph
for finding the centre of gravity of an area about its end ordi
nate fails when applied to such a case as the above. For
instance, try the following example :
A curve has ordinates 10, 9, 7 feet long, 4 feet apart. To
find the position of the centre of gravity of the portion between
the two first ordinates with respect to the end ordinate.
Ordinates.
Simpson's
multipliers.
Functions.
Multipliers
for moment.
Products for
moment.
IO
5
5
O
9
8
72
I
72
7
i
7
2
I 4
. us 5
Centre of gravity from the end ordinate would be
58 X 4 <
Now this is evidently wrong, since the shape of the curve is
such that the centre of gravity ought to be slightly less than
2 feet from the end ordinate.
We must use the following rule :
To ten times the middle ordinate add three times the near
end ordinate and subtract the far end ordinate. Multiply tht
Theoretical Naval Architecture.
remainder by onetwentyfourth the square of the common interval.
The product will be (he moment about the end ordinate.
Using jj, j/ 2 , y^ for the lengths of the ordinates, and h the
common interval, the moment of the portion between the
ordinates y and y 2 about the ordinate y l is given by
We will now apply this rule to the case considered above.
OrHinafPC
Area.
Moment.
Simpson's
multipliers.
Functions.
Simpson's
multipliers.
Functions.
to
9
I
50
72
3
10
30
9
7
1
7
i
 7
Moment
Area
"3
Therefore distance of the centre of gravity from the end oidi
nate is
"3 X H = IJ 3 X 2 x 3
115 X Yt 115 X 3
= fff =I '965 feet
This result is what one might expect by considering the
shape of the curve.
To find the Position of the Centre of Gravity of a
Curvilinear Area with respect to its Base. Let DABC,
Fig 36, be a plane curvilinear area. We wish to find the
distance of its centre of gravity from the base DC. To do
this, we must first find the moment of the figure about DC and
divide it by the area. Take any ordinate PQ, and at PQ
draw a consecutive ordinate giving a strip A# wide. Then the
area of the strip is
y x A.* very nearly
Moments, Centre of Gravity, Centre of Buoyancy, etc. 59
and regarding it as a rectangle, its centre of gravity is at a
distance of \y from the base. Therefore the moment of the
strip about the base is Q
\f X A*
If now we consider the strip
to be indefinitely thin, its
moment about the base will
be
and the moment of the total D
area about the base must be
the sum of the moments of all such strips, 01
p
FIG. 36.
This expression for the moment is of the same form as that for
the area, viz. jy . dx. Therefore, instead of y we put \y* l
through Simpson's rule in the ordinary way, and the result will
be the moment of the area about DC.
Example. An athwartship coalbunker is 6 feet long in a foreandaft
direction. It is bounded at the sides by two longitudinal bulkheads 34 feet
apart, and by a horizontal line at the top. The bottom is formed by the
inner bottom of the ship, and is in the form of a curve having vertical
ordinates measured from the top of 12*5, 15*0, 16*0, l6'3, 16*4, 16*3, 16*0,
15*0, 1 2* 5 feet respectively, the first and last ordinates being on the bulk
heads. Find
(1) The number of tons of coal the bunker will hold.
(2) The distance of the centre of gravity of the coal from the top.
The inner bottom is symmetrical either side of the middle line, so we
need only deal with one side. The work is arranged as follows ;
Ordinates.
Simpson's
multipliers.
Functions of
ordinates.
Squares of
ordinates.
Simpson's
multipliers.
Functions
of squares.
I6'4
I
164
269
I
269
l6 3
4
6 5 2
266
4
1064
160
2
3 20
2 5 6
2
512
150
4
OO'O
225
4
900
125
I
125
I 5 6
i
I 5 6
Function of area 186*1
Function of moment 2901
1 This assumes that Simpson's first rule, which will most probably be
used, will correctly integrate a parabola of the fourth order, which can be
shown to be the case for all practical purposes.
60 Theoretical Naval Architecture.
Common interval = 4*25 feet
Halfarea of section = l86'i X J X 4*25 square feet
Volume of bunker = iS6'i x 4 ' 2g x 2 x 6 cub i c feet
Number of tons of coal = 186*1 X J
= 72 tons
Moment of halfarea below top = 2901 X  X ^^
And distance of C.G. from the top = 
_
= 78 feet.
In the first three columns we proceed in the ordinary way
for finding the area. In the fourth column is placed, not the
halfsquares, but the squares of the ordinates in column i, the
multiplication by \ being brought in at the end. These
squares are then put through Simpson's multipliers, and the
addition of column 6 will give a function of the moment of
the area about the base. This multiplied by \ and by \ the
common interval gives the actual moment This moment
divided by the area gives the distance of the centre of gravity
we want. It will be noticed that \ the common interval
comes in top and bottom, so that we divide the function of the
moment 2901 by the function of the area 186*1, and then
multiply by \ to get the distance of centre of gravity required.
It is not often required in practice to find the centre of
gravity of an area with respect to its base, because most of the
areas we have to deal with are symmetrical either side of a
centre line (as waterplanes), but the problem sometimes occurs,
the question above being an example.
To find the Position of the Centre of Gravity of
an Area bounded by a Curve and Two Radii. We
have already seen (p. 15) how to find the area of a figure such
as this. It is simply a step further to find the position of the
centre of gravity with reference to either of the bounding radii.
Let OAB, Fig. 13, be a figure bounded by a curve, AB, and
two bounding radii, OA, OB. Take any radius OP, the angle
BOP being called * 0, and the length of OP being called r.
Moments, Centre of Gravity, Centre of Buoyancy,, etc. 6 1
Draw a consecutive radius, OP' ; the angle POP' being indefi
nitely small, we may call it dd. Using the assumptions we
have already employed in finding areas, the area POP' =
Jr 2 .<#?, POP' being regarded as a triangle. The centre of
gravity of POP' is at g t and Og = fr, and gm is drawn perpen
dicular to OB, andgm = \r . sin 6 (see p. 91).
 *>>< **<>
= r*.sin Q.dB
The moment of the whole figure about OB is the sum of
the moments of all such small areas as POP', or, using the
ordinary notation
This is precisely similar in form to the expression we found
for the area of such a figure as the above (see p. 15), viz.
so that, instead of putting ^ through Simpson's rule, measuring
r at equidistant angular intervals, we put \r* . sin 6 through
the rule in a similar way. This will be best illustrated by the
following example :
Example. Find the area and position of centre of gravity of a quadrant
of a circle with reference to one of its bounding radii, the radius being
10 feet.
We will divide the quadrant by radii 15 apart, and thus be able to use
Simpson's first rule.
O .
.
,'d
.
Ij
tt"
* d
R ,
Number
radius
fi
a
I
c .Ji
if
Product
area.
I
H
if
<< 1C
If
l.fl
Ix
*\
.a
fl
Product
momen
I
10
IOO
I
IOO
IOOO
O'O
O
i
O
2
IO
IOO
4
4OO
IOOO
15
0258
258
4
1,032
3
10
IOO
2
200
IOOO
30
0500
500
2
I,OOO
4
10
IOO
4
4OO
IOOO
45
0707
707
4
2,828
10
IO
IOO
IOO
2
4
200
4 00
IOOO
IOOO
60
75
0866
0965
866
965
2
4
1,732
3,860
7
10
IOO
I
IOO
IOOO
90
I '000
IOOO
I
1,000
Function of area 1800
Function of moment 11,452
62 Theoretical Naval Architecture.
The circular measure of 180 = IT = 31416
= ?8'54 square feet
Moment of area about the first radius = 11,452 X X f x  J
therefore distance of centre of gravity from the first radius is
Moment f area =
11452 x 2
= l8cx>x 3 = 424feet
The exact distance of the centre of gravity of a quadrant
from either of its bounding radii is times the radius, and if
3 71 "
this is applied to the above example, it will be found that the
result is correct to two places of decimals, and would have
been more correct if we had put in the values of the sines of
the angles to a larger number of decimal places.
Centre of Gravity of a Solid Body which is
bounded by a Curved Surface and a Plane. In the
first chapter we saw that the finding the volume of such a solid
as this was similar in principle to the finding the area of a
plane curve, the only difference being that we substitute areas
for simple ordinates, and as a result get the volume required.
The operation of finding the centre of gravity of a volume in
relation to one of the dividing planes is precisely similar to the
operation of finding the centre of gravity of a curvilinear area
in relation to one of its ordinates. This will be illustrated by
the following example :
Example. A coalbunker has sections 17* 6" apart, and the areas of
these sections, commencing from forward, are 98, 123, 137, 135, 122 square
feet respectively. Find the volume of the bunker, and the position of its
centre of gravity in a foreandaft direction.
Moments, Centre of Gravity, Centre of Buoyancy, etc. 63
Areas.
Simpson's
multipliers.
Functions of
areas.
Number of
intervals from
forward.
Products for
moments.
9 8
I
9 8
o
123
4
492
I
492
137
2
274
2
548
135
4
540
3
l62O
122
I
122
4
4 88
1526
3148
Volume =1526x^x17$ = 8902 cubic feet
moment = 3148 x \ X 17$ X 17$
/. distance of centre of gravity \ _ 3*48 X I7'5
from forward end / 1526
36*2 feet
It is always advisable to roughly check any result such
as this ; and if this habit is formed, it will often prevent mis
takes being made. The total length of this bunker is 4 x
17' 6" = 70 feet, and the areas of the sections show that the
bunker is fuller aft than forward, and so, on the face of it, we
should expect the position of the centre of gravity to be some
what abaft the middle of the length ; and this is shown to be
so by the result of the calculation. Also as regards the volume.
This must be less than the volume of a solid 70 feet long, and
having a constant section equal to the area of the middle
section of the bunker. The volume of such a body would
be 70 X 137 = 9590 cubic feet. The volume, as found by
the calculation, is 8902 cubic feet, thus giving a coefficient of
IHHrf = '93 nearly, which is a reasonable result to expect.
Centre of Buoyancy. The centre of buoyancy of a
vessel is the centre of gravity of the underwater volume, or,
more simply, the centre of gravity of the displaced water.
This has nothing whatever to do with the centre of gravity of
the ship herself. The centre of buoyancy is determined solely
by the shape of the underwater portion of the ship. The
centre of gravity of the ship is determined by the distribution
of the weights forming the structure, and of all the weights she
has on board. Take the case of two sister ships built from
the same lines, and each carrying the same weight of cargo
and floating at the same waterline. The centre of buoyancy
6 4
Theoretical Naval Architecttire.
of each of these ships must necessarily be in the same position.
But suppose they are engaged in different trades the first,
say, carrying a cargo of steel rails and other heavy weights,
which are stowed low down. The second, we may suppose,
carries a cargo of homogeneous materials, and this has to be
stowed much higher than the cargo in the first vessel. It is
evident that the centre of gravity in the first vessel must be
much lower down than in the second, although as regards form
they are precisely similar. This distinction between the centre
of buoyancy and the centre of gravity is a very important
one, and should always be borne in mind.
To find the Position of the Centre of Buoyancy of
a Vessel in a Foreandaft Direction, having given
the Areas of Equidistant Transverse Sections. The
following example will illustrate the principles involved :
Example. The underwater portion of a vessel is divided by transverse
sections 10 feet apart of the following areas, commencing from forward : 0*2,
227, 488, 732, 884, 828, 587, 262, 39 square feet respectively. Find
the position of the centre of buoyancy relative to the middle section.
Number of
station.
Area of
section.
Simpson's
multipliers.
Functions of
area.
Number of
intervals
from middle.
Product for
moment.
,
0'2
,
0'2
4
08
2
22'7
4
908
3
2724
3
488
2
97'6
a
1952
4
73 2
4
2928
i
2928
5
88'4
2
1768
7612
6
828
4
331'2
i
33i'2
7
587
2
II7'4
2
2348
8
262
4
I048
3
3i4'4
9
3'9
3'9
4
156
Function of displacement 12155
Function of
moment
\ 8 9 6c
Volume of displacement = 12155 X g
excess of products aft = 896*0 761*2 = 134*8
moment aft = 1348 X ^ X 10
C.B. abaft middle = '34* x x 10
I2I55 * 5 Q
12155
Moments, Centre of Gravity, Centre of Buoyancy, etc. 65
The centre of gravity of a plane area is fully determined
when we know its position relative to two lines in the plane,
which are generally taken at right angles to one another. The
centre of gravity of a volume is fully determined when we
know its position relative to three planes, which are generally
taken at right angles to one another. In the case of the under
water volume of a ship, we need only calculate the position of
its centre of gravity relative to (i) the load waterplane, and
(2) an athwartship section (usually the section amidships),
because, the two sides of the ship being identical, the centre
of gravity of the displacement must lie in the middleline
longitudinal plane of the ship.
Approximate Position of the Centre of Buoyancy.
In vessels of ordinary form, it is found that the distance of the
centre of buoyancy below the L.W.L. varies from about ^ to ^
of the mean draught to top of keel, the latter being the case
in vessels of full form. For yachts and vessels of unusual
form, such a rule as this cannot be employed.
Example, A vessel 13' 3" mean draught has her C.B. 5*34 feet below
L.^V.L.
Here the proportion of the draught is
5 '34 8 06
.sLsCL = o'4.O1 =
1325 4 6 20
This is an example of a fine vessel.
Example. A vessel 27' 6" mean draught has her C.B. 12*02 feet below
L.W.L.
Here the proportion of the draught is
1202 __ 875
275 ~ 20
This is an example of a fuller vessel than the first case.
Morrish's Approximate Formula for the Distance
of the Centre of Buoyancy below the Load Water
line. 1
Let V = volume of displacement up to the loadline in
cubic feet ;
A = the area of the load waterplane in square feet ;
d the mean draught (to top of keel) in feet.
1 See a paper in Transactions of the Institution of Naval Architects, by
Mr. S. W. F. Morrish. M.I.N.A., in 1892.
66
Theoretical Naval Architecture.
Then centre of buoyancy below L.W.L. = J (  + %
This rule gives exceedingly good results for vessels of
ordinary form. In the early stages of a design the above
particulars would be known as some of the elements of the
design, and so the vertical position of the centre of buoyancy
can be located very nearly indeed. In cases in which the
stability of the vessel has to be approximated to, it is important
to know where the C.B. is, as will be seen later when we are
dealing with the question of stability.
The rule is based upon a very ingenious assumption, as
follows :
In Fig. 36A, let ABC be the curve of water plane areas, DC
being the mean draught d. Draw the rectangle AFCD. Make
y
DE = r D say. Draw EG parallel to DA cutting the diagonal
*TL
FD in H. Finish the figure as indicated. Then the assumption
A. _ D.
E.
C.
FIG. 36A.
made is that the C.G. of the area DAHC is the same distance below
the waterline as the C.G. of DABC which latter, of course, gives
the distance below the waterline of the centre of buoyancy. It is
seen by inspection that the assumption is a reasonable one.
DAHC and DABC have the same area as we now proceed to
show. The rectangles AH and HC are equal, so that the triangles
AGH and HEC are equal, and therefore
Area of AHCD = area of AGED
Moments, Centre of Gravity, Centre of Buoyancy, etc. 67
The latter gives the volume of displacement, as it is a rectangle
having sides equal to A and r respectively. The area of DABC
also gives the volume of displacement, so that DAHC and DABC
are equal.
We now have to determine the distance of the C.G. of DAHC
below the waterline.
Area AGH \ X AG X GH GH
Area AGED ~ AG X AD & ' AD
d being the draught.
/. Area AGH = \ * rectangle AGED
We may regard the figure DAHC made up by taking away
AGH from the rectangle AE and putting it in the position HEC.
The shift of its C.G. downwards is \.d. Therefore the C.G. of
the whole figure will shift downwards, using the principle explained
in p. 100, a distance x, given by
AGED x x = AGH x 
or putting in the value found above for the area of AGH, we have
The C.G. of AGED is a distance below the waterline. There
2
.fore the C.G. of DAHC is below the waterline, a distance
which is Morrish's approximation to the distance of the C.B. below
the waterline.
The Area of a Curve of Displacement divided by
the Load Displacement gives the Distance of the
Centre of Buoyancy below the Load Waterline. This
is an interesting property of the curve of displacement. The
demonstration is as follows :
Let OBL, Fig. 363, be the curve of displacement of a vessel
constructed in the ordinary way, OW being the mean draught and
WL being the displacement in tons.
Take two level lines AB, A'B', a short distance apart, &z say.
Call the area of the waterplane at the level of AB, A square feet,
68
Theoretical Naval Architecture.
and the distance of this waterplane below the WL, z. The
volume between AB and A'B' is A X As, or supposing they are
indefinitely close together A X dz. The moment of this layer
about the WL is A x dz x (z + \ . dz) = A . z . dz, neglecting the
t C'C.
or
FIG. 368.
square of the small quantity dz. The distance of the C.B. below
WL is the sum of all such moments divided by the displacement
volume,
j&.z.dz
35 x WL
Now the difference between the lengths of A'B' and AB is the
weight of the water between these level lines or % . A . dz. The
area of the whole figure is given by the summatim of all such
areas as the strip B'C, which has a length z and a breadth
^g . Adk. Area of figure is therefore aV/A .z.dz, and this divided
by the displacement is
35 x WL
which is the expression found above for the distance of the C.B.
below the WL.
Example. Draw a curve of displacement for all draughts of a cylindrical
vessel 20 feet diameter and 150 feet long, and find by using the curve the
distance of the C.B. from the base when floating (a) at 10 feet level draught,
(b) at 15 feet level draught.
Ans. (a) 576 feet ; (b) 825 feet.
If a new curve be drawn having for ordinates the area of the curve of
displacement at respective levels, it may readily be shown that the tangent
to this curve at any draught will intersect the scale of draughts at the
height of the centre of buoyancy. This new curve is a curve of moments
Moments, Centre of Gravity, Centre of Buoyancy, etc. 69
of displacement up to each level line about such level line. By construct
ing such a curve in the graphic method of finding displacement (see later),
considerable simplification of the process is obtained. Thus, in Fig. 39,
AHL is the curve of displacement. By integrating this curve (still by the
graphic method), a curve of moments of displacements is obtained, the
ordinate of which on BR will be the moment of displacement BL about
the L.W.L. This moment, divided by the displacement BL, gives the
distance of the C.B. below the L.W.L. This may be checked by drawing
the tangent to the new curve, as seen above. In a similar manner, in
Fig. 40 the difference between the areas of BB in the fore and after bodies
divided by the total displacement gives the fore and aft position of the C.B.
with reference to No. 6 station.
Displacement Sheet. 1 We now proceed to investigate
the method that is very generally employed in practice to find
the displacement of a vessel, and also the position of its centre
of buoyancy both in a longitudinal and a vertical direction.
The calculation is performed on what is termed a " Displace
ment Sheet" or "Displacement Table" and a specimen calcula
tion is given at the end of the book for a singlescrew tug of
the following dimensions :
Length between perpendiculars 74' o'
Breadth moulded 14' 6'
Depth moulded 8' 3'
Draught moulded forward 5' 5'
aft 6' 2'
,, mean 5' 9
The sheer drawing of the vessel is given on Plate I. This
drawing consists of three portions the body plan, the half
breadth plan, and the sheer. The sheer plan shows the ship
in side elevation, the load waterline being horizontal, and the
keel, in this case, sloping down from forward to aft. The ship
is supposed cut by a number of transverse vertical planes,
which are shown in the sheer plan as straight lines, numbered
i, 2, 3, etc. Now, each of these transverse sections of the ship
has a definite shape, and the form of each halfsection to the
outside of frames is shown in the bodyplan, the sections being
numbered as in the sheer. The sections of the forward end
form what is termed the "forebody? and those of the after
end the " afterbody" Again, the ship may be supposed to be
cut by a series of equidistant horizontal planes, of which the
1 For displacement sheet with combination of Simpson's first rule and
Tchebycheffs rule, see Appendix A.
70 Theoretical Naval Architecture.
load waterplane is one. The shape of the curve traced on each
of these planes by the moulded surface of the ship is given in
the halfbreadth plan, and the curves are numbered A, i, 2, 3,
etc., to agree with the corresponding lines in the sheer and
body plan. Each of these plans must agree with the other
two. Take a special station, for example, No. 4. The breadth
of the ship at No. 4 station at the level of No. 3 waterplane is
Oa' in the bodyplan, but it is also given in the halfbreadth plan
by Oa, and therefore Oa must exactly equal Oa'. The process
of making all such points correspond exactly is known as
"fairing? For full information as to the methods adopted in
fairing, the student is referred to the works on " Layingoff"
given below. 1 For purposes of reference, the dimensions of
the vessel and other particulars are placed at the top of the
displacement sheet. The waterlines are arranged on the
sheer drawing with a view to this calculation, and in this case
are spaced at an equidistant spacing apart of i foot, with an
intermediate waterline between Nos. 5 and 6. The number
of waterlines is such that Simpson's first rule can be used, and
the multipliers are, commencing with the load waterplane
i 4 2 4 ji 2 
The close spacing near the bottom is very necessary to
ensure accuracy, as the curvature of sections amidships of the
vessel is very sharp as the bottom is approached, and, as we
saw on p. 13, Simpson's rules cannot accurately deal with areas
such as these unless intermediate ordinates are introduced.
Below No. 6 waterplane there is a volume the depth of which
increases as we go aft, and the sections of this volume are very
nearly triangles. This volume is dealt with separately on the
lefthand side of the table, and is termed an " appendage?
In order to find the volume of displacement between water
planes i and 6, we can first determine the areas of the water
planes, and then put these areas through Simpson's rule. To
find the area of any of the waterplanes, we must proceed in
the ordinary manner and divide its length by ordinates so that
1 " Laying Off," by Mr. S. J. P. Thearle ; " Laying Off," by Mr. T. II.
Watson ; " Laying Off," by Messrs. Alt wood and Cooper.
Moments, Centre of Gravity, Centre of Buoyancy, etc. 7 1
Simpson's rule (preferably the first rule) can be used. In the
case before us, the length is from the afteredge of the stem to
the forward edge of the body post, viz. 71 feet, and this
length is divided into ten equal parts, giving ordinates to each
of the waterplanes at a distance apart of 7*1 feet. The dis
placementsheet is arranged so that we can put the lengths of
the semiordinates of the waterplanes in the columns headed
respectively L.W.L., 2\Y.L., 3"W.L., etc., the semiordinates at
the several stations being placed in the same line as the numbers
of ordinates given at the extreme left of the table. The lengths
of the semiordinates are shown in italics. Thus, for instance,
the lengths of the semiordinates of No. 3 W.L., as measured
off, are 005, 1*82, 4*05, 590, 690, 725, 704, 651 5^
2*85, and 0*05 feet, commencing with the forward ordinate
No. i, and these are put down in italics 1 as shown beneath
the heading 3 W.L. in the table. The columns under the
heading of each W.L. are divided into two, the semiordinates
being placed in the first column. In the second column of each
waterline is placed the product obtained by multiplying the
semiordinate by the corresponding multiplier to find the area.
These multipliers are placed at column 2 at the left, opposite
the numbers of the ordinates. We have, therefore, under the
heading of each waterline what we have termed the "functions
of ordinates" and if these functions are added up, we shall
obtain what we have termed the "function of area"
Taking No. 3 W.L. as an instance, the "function" of its
area is 144*10, and to convert this "function" into the actual
area, we must multiply by onethird the common interval to
complete Simpson's first rule, i.e. by \ X 7*i ; and also by 2
to obtain the area of the waterplane on both sides of the ship.
We should thus obtain the area of No. 3 W.L.
14410 x \ X 71 X 2 = 68907 square feet
The functions of the area of each waterplane are placed at
the bottom of the columns, the figures being, starting with
the L.W.L., 16370, 15536, 14410, 12874, 10567, 8727,
and 6097. To get the actual areas of each of the waterplanes,
1 In practice, it is advisable to put down the lengths of the semi
ordinates in some distinctive colour, such as red.
72 Theoretical Naval Architecture.
we should, as above, multiply each of these functions by \ X
7* i X 2. Having the areas, we could proceed as on p. 26 to
find the volume of displacement between No. i and No. 6
waterlines, but we do not proceed quite in this way; we
put the "functions of areas' 1 through Simpson's rule, and
multiply afterwards by \ X T i X 2, the same result being
obtained with much less work. Below the " functions of
areas " are placed the Simpson's multipliers, and the products
16370, 621*44, etc., are obtained. These products added up
give 1 95 1 "83. This number is a function of the volume of
displacement, this volume being given by first multiplying it
by onethird the vertical interval, i.e. J x i ; and then by
J X 7*1 X 2, as seen above. The volume of displacement,
between No. i W.L. and No. 6 W.L. is therefore
195183 x (J X i) X (J X 7'i) X 2 = 30795 cubic feet
and the displacement in ) 3Q79'5 = 8rg8 tons
tons (salt water) x ) 35
We have thus found the displacement by dividing the
volume under water by a series of equidistant horizontal planes ;
but we could also find the displacement by dividing the under
water volume by a series of equidistant vertical planes, as we
saw in Chapter I. This is done on the displacement sheet,
an excellent check being thus obtained on the accuracy of the
work. Take No. 4 section, for instance : its semiordinates,
commencing with the L.W.L., are 640, 624, 590, 532, 430
340, and 225 feet These ordinates are already put down
opposite No. 4 ordinate. If these are multiplied successively
by the multipliers, i, 4, 2, 4, i, 2, , and the sum of the
functions of ordinates taken, we shall obtain the "function of
area " of No. 4 section between the L.W.L. and 6 W.L. This
is done in the table by placing the functions of ordinates
immediately below the corresponding ordinate, the multiplier
being given at the head of each column. We thus obtain a
series of horizontal rows, and these rows are added up, the
results being placed in the column headed u Function of areas?
Each of these functions multiplied by onethird the common
1 Thirty five cubic feet of salt water taken to weigh one ton.
Moments, Centre of Gravity, Centre of Buoyancy, etc. 73
interval, i.e. \ X i, and then by 2 for both sides, will give
the areas of the transverse sections between the L.W.L. and
6 W.L. ; but, as before, this multiplication is left till the end
of the calculation. These functions of areas are put through
Simpson's multipliers, the products being placed in the column
headed " Multiples of areas''' This column is added up, giving
the result 195183. To obtain the volume of displacement,
we multiply this by (J X i) X 2 X ( X 7'i). It will be noticed
that we obtain the number 195183 by using the horizontal
waterlines and the vertical sections ; and this must evidently
be the case, because the displacement by either method must
be the same. The correspondence of these additions forms
the check, spoken of above, of the accuracy of the work. We
thus have the result that the volume of displacement from
L.W.L. to 6 W.L. is 30795 cubic feet, and the displacement
in tons of this portion 8798 tons in salt water. This is termed
the " Main solid," and forms by far the greater portion of the
displacement.
We now have to consider the portion we have left out
below No. 6 waterplane. Such a volume as this is termed an
" appendage? The sections of this appendage are given in the
bodyplan at the several stations. The form of these sections
are traced off, and by the ordinary rules their areas are found
in square feet. We have, therefore, this volume divided by a
series of equidistant planes the same as the main solid, and we
can put the areas of the sections through Simpson's rule and
obtain the volume. This calculation is done on the lefthand
side of the sheet, the areas being placed in column 3, and the
functions of the areas in column 4. The addition of these
functions is 49*99, and this multiplied by \ x 7'i gives the
volume of the appendage in cubic feet, viz. 1183; an d this
volume divided by 35 gives the number of tons the appendage
displaces in salt water, viz. 338 tons. The total displacement
is thus obtained by adding together the, main solid and the
appendage, giving 9136 tons in salt water. The displacement
in fresh water (36 cubic feet to the ton) would be 888 tons.
The sheer drawing for this vessel as given on Plate I. was
drawn to the frame line, i.e. to the moulded dimensions of the
74 Theoretical Naval Architecture.
ship ; but the actual ship is fuller than this, because of the outer
bottom plating, and this plating will contribute a small amount
to the displacement, but this is often neglected. Some sheer
drawings, on the other hand, are drawn so that the lines include
a mean thickness of plating outside the frame line, and when
this is the case, the displacement sheet gives the actual dis
placement, including the effect of the plating. For a sheathed
ship this is also true; in this latter case, the displacement
given by the sheathing would be too great to be neglected.
When the sheer drawing is drawn to the outside of sheathing,
or to a mean thickness of plating, it is evident that the ship
must be laid off on the mould loft floor, so that, when built,
she shall have the form given by the sheer drawing.
We now have to find the position of the centre of buoyancy
both in a foreandaft and in a vertical direction. (It must be
in the middleline plane of the ship, since both sides are sym
metrical.) Take first the foreandaft position. This is found
with reference to No. 6 station. The functions of the areas of
the sections are 05, 23055, etc., and in the column headed
" Multiples of areas " we have these functions put through
Simpson's multipliers. We now multiply these multiples by
the number of intervals they respectively are from No. 6 station,
viz. 5, 4, etc., and thus obtain a column headed " Moments."
This column is added up for the fore body, giving 150543, and
for the after body, giving 191302, the difference being 40759
in favour of the after body. To get the actual moment of the
volume abaft No. 6 station, we should multiply this difference
by (^ X i) for the vertical direction, (J x 71) for the foreand
aft direction, and by 2 for both sides, and then by 7*1, since we
have only multiplied by the number of intervals away, and not
by the actual distances, or 47'59 X (J x i) X (J X 7*1) x 2
X 7' i. The volume, as we have seen above, is given by
1951*83 x (\ x i) x (i x 71) x 2
The distance of the centre of gravity of the main solid from
No. 6 station will be
Moment  1  volume
Moments, Centre of Gravity, Centre of Buoyancy, etc. 75
But on putting this down we shall see that we can cancel out,
leaving us with
W59.XTI 8feet
195183
which is the distance of the centre of gravity of the main solid
abaft No. 6 station. The distance of the centre of gravity of
the appendage abaft No. 6 station is 4'o feet ; the working is
shown on the lefthand side of the table, and requires no further
explanation. 1 These results for the main solid and for the ap
pendages are combined together at the bottom ; the displacement
of each in tons is multiplied by the distance of its centre of gravity
abaft No. 6 station, giving the moments. The total moment
is 143*73, and the total displacement is 91*36 tons, and this
gives the centre of gravity of the total displacement, or what we
term the centre of 'buoyancy , C.B., 1*57 feet abaft No. 6 station.
Now we have to consider the vertical position of the
C.B., and this is determined with reference to the load water
line. For the main solid the process is precisely similar
to that adopted for finding the horizontal position, with the
exception that we take our moments all below the load water
plane, the number of intervals being small compared with the
horizontal intervals. We obtain, as indicated on the sheet, the
centre of gravity of the main solid at a distance of 2*21 feet
below the L.W.L. For the appendage, we proceed as shown
on the lefthand side of the sheet. When finding the areas of
the sections of the appendage, we spot off as nearly as possible
the centre of gravity of each section, and measure its distance be
low No. 6 W.L. If the sections happen to be triangles, this will,
of course, be onethird the depth. These distances are placed in
a column as shown, and the " functions of areas " are respec
tively multiplied by them, e.g. for No. 4 station the function of
the area is 5*92, and this is multiplied by 0*22, the distance
of the centre of gravity of the section of the appendage below
No. 6 W.L. We thus obtain a column which, added up, gives
a total of 13*78. To get the actual moment, we only have to
1 For the vertical C.G. of the appendage Morrish's rule gives a good
approximation.
76 Theoretical Naval Architecture.
multiply this by \ x yi. The volume of the appendage is
4999 X (5 X 7'i). So that the distance of the centre of
gravity of the whole appendage below No. 6 W.L. is given
by moment T volume, or l^J = 027 feet, and therefore the
centre of gravity of the appendage is 527 feet below the
L.W.L. The results for the main solid and for the appendage
are combined together in the table at the bottom, giving the
final position of the C.B. of the whole displacement as 2*32
feet below the L.W.L.
It will be of interest at this stage to test the two approxi
mations that were given on p. 65 for the distance of the C.B.
below the L.W.L. The first was that this distance would be
from ~ to Q of the mean draught to top of keel (i.e. the mean
moulded draught). For this vessel the distance is 2^32 feet,
and the mean moulded draught is 5' 9^", or 5*8 feet, and so
we have the ratio , or exactly ~. The second approxi
5' 8
mation (Morrish's), p. 65, was
(M)
All these are readily obtainable from the displacement sheet,
and if worked out its value is found to be 2*29 feet. This
agrees fairly well with the actual result, 2*32 feet, the error
being 3 in 232, or less than ij per cent.
For large vessels a precisely similar displacementsheet is
prepared, but it is usual to add in the effect of other appen
dages besides that below the lowest W.L. Specimen calcula
tions are given on Tables II. and Ilia, at the end of the book.
In the former the ordinates are to a mean thickness of
plating. In the latter the moulded surface is used, and the
displacement of the shell plating added as an appendage, being
obtained by Denny's formula given on page 86.
Graphic or Geometrical Method of calculating
Displacement and Position of Centre of Buoyancy.
There is one property of the curve, known as the "parabola of
the second order" (see p. 6), that can be used in calculating
Moments, Centre of Gravity, Centre of Buoyancy, etc. 77
by a graphic method the area of a figure bounded by such a
curve. Let BFC, Fig. 37, be a curve bounding the figure
ABCD, and suppose the curve is a "parabola of the second
order" Draw the ordinate EF
midway between AB and DC ; then
the following is a property of the
curve BFC : the area of the seg
ment BCF is given by two thirds
the product of the deflection GF
and the base AD, or
Area BCF = f X GF x AD
Make GH = GF. Then
Area BCF = GH X AD
Now, the area of the trapezoid ABCD is given by
AD x EG, and consequently
The area ADCFB = AD x EH
Thus, if we have a long curvilinear, we can' divide it up
as for Simpson's first rule, and set off on each of the inter
mediate ordinates twothirds the deflection of the curve above
or below the straight line joining the extremities of the dividing
ordinates. Then add together on a strip of paper all such
distances as EH right along, and the sum multiplied by the
1 This property may be used to prove the rule known as Simpson's first
rule. Call AB, EF, DC respectively y y z , y 3 . Then we have
E.
FIG. 37.
EG
andFG=j 2  EG
EH = EG + GH
^'tZJ'LZ.
3
and calling AE = h, we have
Area ADCFB = fo + 4J 2 +J' 3 )
which is the same expression as given by Simpson's first rule.
78 Theoretical Naval Architecture.
distance apart of the dividing ordinates, as AD, will give the
area required. Thus in Fig. 38, AB is divided into equal parts
FIG
as shown. D and E are joined, also E and C ; MO is set
off = HM, and NP is set off = f NK. Then
Area ADEF = AF X GO
and area FECB = FB X LP
and the whole area ABCD = AF X (GO + LP)
We can represent the area ABCD by a length equal to
GO + LP on a convenient scale, if we remember that this length
has to be multiplied by AF to get the area. This principle can
be extended to finding the areas of longer figures, such as
waterplanes, and we now proceed to show how the displacement
and centre of buoyancy of a ship can be determined by its use.
The assumption we made at starting is supposed to hold good
with all the curves we have to deal, i.e. that the portions
between the ordinates are supposed to be "parabolas of the
second order" This is also the assumption we make when
using Simpson's first rule for finding displacement in the ordi
nary way.
Plate I. represents the ordinary sheer drawing of a vessel,
and the underwater portion is divided by the level waterplanes
shown by the halfbreadth plan. The areas of each of these
planes can be determined graphically as above described, the
area being represented by a certain length obtained by the
addition of all such lengths as GO, etc., Fig. 38, the interval
Moments, Centre of Gravity, Centre of Buoyancy, etc. 79
being constant for all the waterplanes. Let AB, Fig. 39, be
set vertically to represent the extreme moulded draught of the
vessel. Draw BC at right angles to AB, to represent on a
convenient scale the area of the L.W.L. obtained as above.
Similarly, DE, FG are set out to represent on the same scale
the areas of waterplanes 2 and 3, and so on for each water
plane. A curve drawn through all such points as C, E, and G
L.W.I
2.W.L.
FIG. 39.
will give a " curve of areas of waferplanes" Now, the area of this
curve up to the L.W.L. gives us the volume of displacement up
to the L.W.L., as we have seen in Chapter I., and we can readily
find the area of the figure ABCEG by the graphic method, and
this area will give us the displacement up to the L.W.L.
Similarly, the area of ADEG will give the displacement up to
2 W.L., and so on. Therefore set off BL to represent on a
8o Theoretical Naval Architecture.
convenient scale the area of the figure ABCE, DK on the
same scale to represent the area ADEG, and so on. Then
a curve drawn through all such points as L, K will give us a
" curve of displacement" and the ordinate of this curve at any
draught will give the displacement at that draught, BL being
the load displacement.
We now have to determine the distance of the centre
of buoyancy below the L.W.L., and to find this we must get
the moment of the displacement about the L.W.L. and
divide this by the volume of displacement below the L.W.L.
We now construct a curve, BPMA, such that the ordinate at
any draught represents the area of the waterplane at that
draught multiplied by the depth of the waterplane below the
L.W.L. Thus DP represents on a convenient scale the area
of No. 2 waterplane multiplied by DB, the distance below the
L.W.L. The ordinate of this curve at the L.W.L. must evi
dently be zero. This curve is a curve of " moments of areas
of waterplanes" about the L.W.L. The area of this curve up
to the L.W.L. will evidently be the moment of the load dis
placement about the L.W.L., and thus the length BR is set out
to equal on a convenient scale the area of BPMA. Similarly,
DS is set out to represent, on the same scale, the area of
DPMA, and thus the moment of the displacement up to 2 W.L.
about the L.W.L. These areas are found graphically as in the
preceding cases. Thus a curve RSTA can be drawn in, and
BR ' BL, or moment of load displacement about L.W.L. 7
load displacement, gives us the depth of the centre of buoyancy
for the load displacement below the L.W.L.
Exactly the same course is pursued for finding the displace
ment and the longitudinal position of the centre of buoyancy,
only in this case we use a curve of areas of transverse sections
instead of a curve of areas of waterplanes, and we get the
moments of the transverse areas about the middle ordinate.
Fig. 40 gives the forms the various curves take for the fore
body. AA is the " curve of areas of transverse sections ; " BB
is the " curve of displacement " for the fore body, OB being the
displacement of the fore body. CC is the curve of " moments
of areas of transverse sections " about No. 6 ordinate ; DD is
Moments, Centre of Gravity, Centre of Buoyancy, etc. 8 1
the curve of "moment of displacement" about No. 6 ordinate,
OD being the moment of the forebody displacement about
No. 6 ordinate. Similar curves can be drawn for the after
body, and the dif
ference of the mo
ments of the fore and
after bodies divided
by the load displace
ment will give the
distance of the centre
of buoyancy forward
or aft of No. 6 ordi
nate, as the case may be. The total displacement must be the
same as found by the preceding method. 1
Method of finding Areas by Means of the Plani
meter. This instrument is frequently employed to find the
area of plane curvilinear figures, and thus the volume of dis
placement of a vessel can be determined. One form of the
instrument is shown in diagram by Fig. 41. It is supported at
three places : first, by a weighted pin, which is fixed in position
by being pressed into the paper ; second, by a wheel, which
actuates a circular horizontal disc, the wheel and disc both
being graduated ; and third, by a blunt pointer. The instru
ment is placed on the drawing, the pin is fixed in a convenient
position, and the pointer is placed on a marked spot A on the
boundary of the curve of which the area is required. The
reading given by the wheel and disc is noted. On passing
round the boundary of the area with the pointer (the same way
as the hands of a clock) back to the startingpoint, another
reading is obtained. The difference of the two readings is
proportional to the area of the figure, the multiplier required to
convert the difference into the area depending on the instru
ment and on the scale to which the figure is drawn. Particu
lars concerning the necessary multipliers are given with the
instrument ; but it is a good practice to pass round figures of
known area to get accustomed to its use.
1 This method may be considerably simplified by using the property
of the curve of displacement given on p. 66.
G
82 Theoretical Naval Architecture.
By the use of the planimeter the volume of displacement of
a vessel can very readily be determined. The body plan is
taken, and the L.W.L. is marked on. The pointer of the in
strument is then passed round each section in turn, up to the
L.W.L., the readings being tabulated. If the differences of the
readings were each multiplied by the proper multiplier, we
should obtain the area of each of the transverse sections, and
so, by direct application of Simpson's rules, we should find the
POINTER.
FIG.
required volume of displacement. Or we could put the actual
difference of readings through Simpson's multipliers, and
multiply at the end by the constant multiplier.
It is frequently the practice to shorten the process as
follows : The bodyplan is arranged so that Simpson's first rule
will be used, i.e. an odd number of sections is employed.
The pointer is passed round the first and last sections, and
the reading is recorded. It is then passed round all the even
sections, 2, 4, 6, etc., and the reading is recorded. Finally,
it is passed round all the odd sections except the first and last,
viz. 3, 5> 7> et c., and the reading is put down. The differences
of the readings are found and put down in a column. The
Moments, Centre of Gravity, Centre of Buoyancy, etc. 83
first difference is multiplied by i, the next difference is multi
plied by 4, and the last by 2. The sum of these products is
then multiplied for Simpson's first rule, and then by the proper
multiplier for the instrument and scale used. The work can
conveniently be arranged thus :
Numbers of
sections.
Readings.
Differences
of readings.
Simpson's
multipliers.
Products.
Initial reading
5 I2 4
__
__
I, 21
S.S^O
2 3 6
I
236
2, 4,6, 8, 10, )
12, 14, 16, [
18,681
13,321
4
53,284
18, 20 ... )
3, 5> 7, 9, ii, 1
13, 15, !7 [
31,758
13,077
2
26,154
19 )
79,674
The multiplier for the instrument and scale of the drawing
used and to complete the use of Simpson's first rule is ; so
that the volume of displacement is 79,674 X cubic feet, and
the displacement in tons is 79,674 x X A = 2732 tons.
There are two things to be noticed in the use of the plani
meter: first, it is not necessary to set the instrument to the
exact zero, which is somewhat troublesome to do ; and second,
the horizontal disc must be watched to see how many times
it makes the complete revolution, the complete revolution
meaning a reading of 10,000.
It is also possible to find the vertical position of the centre
of buoyancy by means of the planimeter. By the method
above described we can determine the displacement up to each
waterline in succession, and so draw in on a convenient scale
the ordinary curve of displacement. Now we can run round
this curve with the planimeter and find its area. This area
divided by the top ordinate (i.e. the load displacement) will
give the distance of the centre of buoyancy below the loadline
(see p. 67).
To find the centre of buoyancy in a foreandaft direction,
it is necessary to tabulate the differences for each section,
and treat these differences in precisely the same way as the
Theoretical Naval Architecture.
" functions of areas of vertical sections " are treated in the
ordinary displacement sheet.
Method of approximating to the Area of the
Wetted Surface 1 by "Kirk's" Analysis. The ship is
assumed to be represented by a block model, shaped as
shown in Fig. 42, formed of a parallel middle body and a
r
FIG. 48.
tapered entrance and run which are taken as of equal length.
The depth of the model is equal to the mean draught, and the
length of the model is equal to the length of the vessel. The
breadth is not equal to the breadth of the vessel, but is equal
to area of immersed midship section f mean draught. The
displacement of the model is made equal to that of the vessel.
We then have
Volume of displace] _ y s
ment )
= AG X area of midship section
V
~ area of midship section
.'. length of entrance} . V
> = length Of Ship  c r^r.
or run area of midship section
V
B' X D
1 The area of wetted surface can be closely approximated to by putting
a. curve of girths (modified for the slope of the level lines, see p. 228)
through Simpson's rule.
Moments. Centre of Gravity, Centre of Buoyancy, etc. 85
where L = length of ship ;
B' = breadth of model ;
D = mean draught.
Having found these particulars, the surface of the model
can be readily calculated.
Area of bottom = AG X B'
Area of both sides = 2(GH f 2AE) x mean draught
The surface of a model formed in this way approximates
very closely to the actual wetted surface of the vessel. It is
stated that in very fine ships the surface of the model exceeds
the actual wetted surface by about 8 per cent., for ordinary
steamers by about 3 per cent., and for full ships by 2 per cent.
By considering the above method, we may obtain an
approximate formula for the wetted surface
V
Area of bottom = ~
Area of sides = 2L/D
where L' is the length along ADCB. Then
V
Surface = 2L'D f ^
This gives rather too great a result, as seen above ; and if
we take
V
Surface = 2LD + ^
we shall get the area of the wetted surface slightly in excess,
but this will allow for appendages, such as keels, etc.
Since V = k . LED, where k is the block coefficient of
displacement, we may write
. Surface = 2LD + k . LB
Approximate Formulae for finding Wetted Surface.
Mr. Denny gives the following formula for the area of
wetted surface :
86
Theoretical Naval Architecture.
i 7 LD + ^
which is seen to be very nearly that obtained above.
Mr. Taylor, in his work on " Resistance and Propulsion
of Ships," gives the following formula :
155 v/WL
where W is the displacement in tons.
The following formula for the wetted surface is used at the
experimental tank at Haslar
One of these formulae can be used to find the area of
wetted surface to calculate the displacement of the skin plating,
as is necessary when the sheer drawing is drawn to the
moulded surface of the ship. See Brown's Displacement Sheet
in Appendix, in which Denny's approximation is used.
EXAMPLES TO CHAPTER IL
I. A ship has the following weights placed on board :
20 tons
45
15
60
40
30
100 feet before amidships
80
40 ,,
50 feet abaft ,,
80
no
Show that these weights will have the same effect on t. r m.u of the ship
as a single weight of 210 tons placed 15$ feet abaft amidships.
2. Six weights are placed on a drawingboard. The weights are 3, 4,
5, 6, 7, 8 Ibs. respectively. Their distances from one edge are 5, 4, 4, 3},
3, 2 feet respectively, and from the edge at right angles, &, \, I, i$, 2, 2$,
feet respectively. The drawingboard weighs 6 Ibs., and is 6 feet long
and 3 feet broad. Find the position where a single support would need
to be placed in order that the board should remain horizontal.
Ans. 3*27 feet from short edge, 158 feet from long edge.
Moments, Centre of Gravity \ Centre of Buoyancy, etc. 87
3. An area bounded by a curve and a straight line is divided by ordinates
4 feet apart of the following lengths : o, 12*5, 143, 15*1, 155, 154, 14*8,
14*0, o feet respectively. Find
(1) Area in square feet.
(2) Position of centre of gravity relative to the first ordinate.
(3) Position of the centre of gravity relative to the base.
Ans. (i) 423 square feet ; (2) 16*27 feet ; (3) 7*24 feet.
4. A triangle ABC has its base EC 15 feet long, and its height 25
feet. A line is drawn 10 feet from A parallel to the base, meeting AB and
AC in D and E. Find the distance of the centre of gravity of DBCE
from the apex.
Ans. i8'57 feet.
5. The semi ordinates of a waterplane in feet, commencing from the
after end, are 5'2, ICT2, 144, 179, 2O'6, 227, 243. 255, 26*2, 26*5,
266, 263, 254, 239, 2i'8, i8'8, 154, 115, 72, 33, 22. The distance
apart is 15 feet. Find the area of the water plane, and the position of the
centre of gravity in relation to the middle ordinate.
Ans. 11,176 square feet; 1015 feet abaft middle.
6. Find the area and transverse position of the centre of gravity of
"half" a waterline plane, the ordinates in feet being 0*5, 6, 12, 16, 12, 10,
and 0*5 respectively, the common interval being 15 feet.
Ans. 885 square feet ; 6*05 feet.
7. The areas of sections 17' 6" apart through a bunker, commencing
from forward, are 65, 98, 123, 137, 135, 122, 96 square feet respectively.
The length of bunker is 100 feet, and its fore end is i' 6" forward of the
section whose area is 65 square feet. Draw in a curve of sectional areas,
and obtain, by using convenient ordinates, the number of cubic feet in the
bunker, and the number of tons of coal it will contain, assuming that 43 cubic
feet of coal weigh i ton. Find also the position of the C.G. of the coal
relative to the after end of the bunker.
Ans. 272 tons ; 46 J feet from the after end.
8. The tons per inch in salt water of a vessel at waterlines 3 feet
apart, commencing with the L.W.L., are 31*2, 3o  o, 2835, 2 6'2i, 23*38,
195, I2'9. Find the displacement in salt and fresh water and the position
of the C.B. below the L.W.L., neglecting the portion below the lowest
W.L. Draw in the tons per inch curve for salt water to a convenient scale,
and estimate from it the weight necessary to be taken out in order to lighten
the vessel 2' 3^" from the L.W.L. The mean draught is 20' 6".
Ans. 5405 tons ; 5255 tons ; 8'OI feet ; 847 tons.
9. In the preceding question, calling the L.W.L. i, find the displacement
up to 2 W.L., 3 W.L., and 4 W.L., and draw in a curve of displacement
from the results you obtain, and check your answer to the latter part of the
question.
10. The tons per inch of a ship's displacement at waterlines 4 feet
apart, commencing at the L.W.L., are 44*3, 427, 40' 5, 37 5, 33'3 Find
number of tons displacement, and the depth of C.B. below the top W.L.
Ans. 7670 tons ; 7 '6 feet.
11. The ship in the previous question has two watertight transverse
bulkheads 38 feet apart amidship, and watertight flats at 4 feet below and
3 feet above the normal L.W.L. If a hole is made in the side 2 feet
below the L.W.L., how much would the vessel sink, taking the breadth
of the L.W.L. amidships as 70 feet? Indicate the steps where, owing to
insufficient information, you are unable to obtain a perfectly accurate result.
Ans. 8 inches.
12. The areas of transverse sections of a coal bunker 19 feet apart are
88 Theoretical Naval Architecture.
respectively 632, 93*6, 121 '6, io8'8, 948 square feet, and the centres
of gravity of these sections are ID'S, ir6, I2'2, 117, 11*2 feet respectively
below the L.W.L. Find the number of tons of coal the bunker will hold,
and the vertical position of its centre of gravity (44 cubic feet of coal to the
ton).
Ans. 1743 tons; ii'68 feet below L.W.L.
13. A vessel is 180 feet long, and the transverse sections from the load
waterline to the keel are semicircles. Find the longitudinal position of
the centre of buoyancy, tke ordinates of the load waterplane being I, 5, 13,
15, 14, 12, and 10 feet respectively.
Ans, 1 06 '2 feet from the finer end.
14. Estimate the distance of the centre of buoyancy of a vessel below
the L.W.L., the vessel having 22' 6" mean moulded draught, block co
efficient of displacement 0x55, coefficient of fineness of L.W.L. O'7 (use
Morrish's formula, p. 65).
Ans. 965 feet.
15. A vessel of 2210 tons displacement, 13' 6" draught, and area
of load waterplane 8160 square feet, has the C.B., calculated on the dis
placement sheet, at a distance of 5*43 feet below the L.W.L. Check this
result.
1 6. The main portion of the displacement of a vessel has been calculated
and found to be 10,466 tons, and its centre of gravity is 1048 feet below
the L.W.L., and 585 feet abaft the middle ordinate. In addition to this,
there are the following appendages :
tons.
Below lowest W.L. 263, 24*8 ft. below L.W.L., 44 ft. abaft mid. ord.
Forward 5, I2'o 202 ft. forward of mid
ord.
Stern ... ... 16, 2'8 ,, 201 ft. abaft mid. ord.
Rudder 16, 175 200 ,,
Bilge keels ... 20, 20 o ,,
Shafting, etc. ... 18, 15 ,, ,, 140 ,, ,,
Find the total displacement and position of the centre of buoyancy.
Ans. 10,804 tons ; C.B. 6'5 abaft mid. ord., IO'86 ft. below L.W.L.
17. The displacements of a vessel up to waterplanes 4 feet apart
are 10,804, 8612, 6511, 4550, 2810, 1331, and 263 tons respectively. The
draught is 26 feet. Find the distance of the centre of buoyancy below the
load waterline.
Ans. ro'9 feet nearly.
18. The load displacement of a ship is 5000 tons, and the centre of
buoyancy is 10 feet below the load waterline. In the light condition the
displacement of the ship is 2000 tons, and the centre of gravity of the layer
between the load and light lines is 6 feet below the loadline. Find the
vertical position of the centre of buoyancy below the light line in the light
condition.
Ans. 4 feet, assuming that the C.G. of the layer is at half its depth.
19. Ascertain the displacement and position of the centre of buoyancy
of a floating body of length 140 feet, depth 10 feet, the forward section
being a triangle 10 feet wide at the deck and with its apex at the keel, and
the after section a trapezoid 20 feet wide at the deck and 10 feet wide at
the keel, the sides of the vessel being plane surfaces; draught of water
may be taken as 7 feet.
Ans. 238 tons ; 56*3 feet before after end, 3 feet below waterline.
20. Show by experiment or otherwise that the centre of gravity of 8
Moments, Centre of Gravity, Centre of Buoyancy, etc. 89
quadrant of a circle 3 inches radius is I '8 inches from the right angle of
the quadrant.
21. A floating body has a constant triangular section, vertex down
wards, and has a constant draught of 12 feet in fresh water, the breadth at
the waterline being 24 feet. The keel just touches a quantity of mud of
specific gravity 2. The waterlevel now falls 6 feet. How far will the
body sink into the mud ?
Ans. 4 feet 11^ inches. *
22. Show that the C.G. of a trapezoid, as ABCD, Fig. 5, is distant
I\~L.) from the middle of the tength h. The C.G. must be on a line
joining the centres of the parallel sides. Thus the position of the C.G.
is fully determined. This will also apply to a figure in which the parallel
sides are not perpendicular to one of the other sides.
23. Apply Morrish's rule to find the C.G. of a semicircle, and state the
error involved if the semicircle is 20 feet radius.
Ans. 0*19 foot.
24. For the lower appendage of the ship displacement sheet, Table I.,
find the vertical C.G. by using Morrish's rule.
25. Describe the process of finding the area and position of the C.G.
of a plane figure by radial integration, and apply it to find these in the case
of a rectangle 8 feet wide and 1 2 feet long.
26. A vessel is 300 feet x 36$ teet X 13^ feet draught, 2135 tons dis
placement. Find the area of wetted surface by each of the formula
given on p. 86.
Ans., Denny, 12,421 square feet.
Taylor, 12,400 square feet.
Froude, 12,350 square feet.
The student is advised to take a sheer drawing and obtain a close
approximation to the wetted surface by putting the half girths to water
line through Simpson's rule. Then to apply the above formulae and see
what the comparative results are.
27. Show, by means of the result in question 22, that the C G. of a
trapezoid in relation to the parallel sides is given by the construction of
Fig. 126.
28. Having given the mean ordinate/of a trapezoid and the distance
x of its C.G. from the larger end, show that the lengths of the parallel
sides are
/ 6Ar\ ,
(4/f)and
where h is the length.
1 This example is worked out at the end of Appendix A.
CHAPTER TIL
CONDITIONS OF EQUILIBRIUM, TRANSVERSE MET A*
CENTRE, MOMENT OF INERTIA, TRANSVERSE BM,
INCLINING EXPERIMENT, METACENTRIC HEIGHT,
ETC.
Trigonometry. The student of this subject will find it a
( distinct advantage, especially when dealing with the question
of stability, if he has a knowledge of some of the elementary
portions of trigonometry. The following are some properties
which should be thoroughly grasped :
Circular Measure of Angles. The degree is the unit gene
rally employed for the measurement of angles. A right angle
is divided into 90 equal
parts, and each of these
parts is termed a "de
gree." If two lines, as
OA, OB, Fig. 43, are
inclined to each other,
forming the angle AOB,
and we draw at any radius
OA an arc AB from the
centre O, cutting OA,
OB in A and B, then
O. A.
Ftc. 43
length of arc AB f radius OA is termed the circular measure
of the angle AOB. Or, putting it more shortly
arc
Circular measure = 
The circular measure of four right )
angles, or 360 degrees
radius
circumference of a circle
radius
27T
Conditions of Equilibrium, Transverse Metacentre, etc.
The circular measure of a right angle / =
Since 360 degrees = 2?r in circular measure, then the angle
whose circular measure is unity is
360
= 57'3 degrees
The circular measure of i degree is 7 = 0*01745, and
thus the circular measure of any angle is found by multiplying
the number of degrees in it by
Trigonometrical Ratios?
etc. Let BOC, Fig. 44, be
any angle ; take any point P
in one of the sides OC, and
draw PM perpendicular to
OB. Call the angle BOC, 6>. 2
PM is termed the perpen
dicular.
OM is termed the base.
OP is termed the hypo
tenuse. O.
Then
IA
BASE.
FIG. 44.
PM perpendicular
7^ =  1  = sine 0, usually written sin 6
OP hypotenuse
OM base
7 ~ = . = cosine 6, usually written cos 6
OP hypotenuse
PM perpendicular . ,.
= C r  = tangent 6, usually written tan 6
These ratios will have the same value wherever P is taken
on the line OC.
1 An aid to memory which is found of assistance by many in learning
these ratios is
Sin /<rrplexes ^y/ocrites t
Cos of base /y/ocrisy.
9 6 is a Greek letter (theta) often used to denote an angle.
9 2
Theoretical Naval Architecture
We can write sin =
cos =
and also tan 6 =
hyp
base
hyp.
sin
cos e
There are names for the inversions of the above ratios,
which it is not proposed to use in this work.
For small angles, the value of the angle in circular
measure is very nearly the same as the values of sin and
tan &. This will be seen by comparing the values of 0, sin 0,
and tan for the following angles :
Angle in
degrees.
Angle in
circular
measure.
Sin A
,Tan 6.
2
0'0349
00349
00349
4
00698
00697
00699
6
01047
01045
01051
8
01396
01392
0I405
1C
01745
01736
01763
Up to 10 they have the same values to two places of
decimals, and for smaller angles the agreement in value is
closer still.
Tables of sines, cosines, and tangents of angles up to 90
are given in Appendix B.
Conditions that must hold in the Case of a Vessel
floating freely, and at Rest in Still Water. We
saw in Chapter I. that, for a vessel floating in still water,
the weight of the ship with everything she has on board must
equal the weight of the displaced water. To demonstrate this,
we imagined the cavity left by the ship when lifted out of the
water to be filled with water (see Fig. 17). Now, the upward
support of the surrounding water must exactly balance the weight
of the water poured in. This weight may be regarded as acting
downwards through its centre of gravity, or, as we now term
it, the centre of buoyancy. Consequently, the upward support
Conditions of Equilibrium, Transverse Metacentre, etc. 93
of the water, or the buoyancy, must act through the centre of
buoyancy. All the horizontal pressures of the water on the
surface of the ship must evidently balance among themselves.
We therefore have the following forces acting upon the ship :
(1) The weight acting downwards through the C.G. ;
(2) The upward support of the water, or, as it is termed,
the buoyancy, acting upwards through the C.B. ;
and for the ship to be at rest, these two forces must act in the
same line and counteract each other. Consequently, we also
have the following condition :
The centre of gravity of the ship, with everything she has on
board, must be in the same vertical line as the centre of buoyancy.
If a rope is pulled at both ends by two men exerting the
same strength, the rope will evidently remain stationary; and
this is the case with a ship floating freely and at rest in still
water. She will have no tendency to move of herself so long
as the C.G. and the C.B. are in the same vertical line.
Definition of Statical Stability. The statical
stability of a vessel is the tendency she has to return to the
upright when inclined away from that position. It is evident
that under ordinary conditions of service a vessel cannot
always remain upright ; she is continually being forced away
from the upright by external forces, such as the action of
the wind and the waves. It is very important that the ship
shall have such qualities that these inclinations that are forced
upon her shall not affect her safety ; and it is the object of the
present chapter to discuss how these qualities can be secured
and made the subject of calculation so far as small angles of
inclination are concerned.
A ship is said to be in stable equilibrium for a given direc
tion of inclination if, on being slightly inclined in that direction
from her position of rest, she tends to return to that position.
A ship is said to be in unstable equilibrium for a given
direction of inclination if, on being slightly inclined in that
direction from her position of rest, she tends to move away
farther from that position.
A ship is said to be in neutral or indifferent equilibrium
for a given direction of inclination if, on being slightly inclined
94
Ttieoretical Naval Architecture.
in that direction from her position of rest, she neither tends
to return to nor move farther from that position.
These three cases are represented by the case of a heavy
sphere placed upon a horizontal table.
1. If the sphere is weighted so that its C.G. is out of the
centre, and the C.G. is vertically below the centre, it will be
in stable equilibrium.
2. If the same sphere is placed so that its C.G. is vertically
above the centre, it will be in unstable equilibrium.
3. If the sphere is formed of homogeneous material so that
its C.G. is at the centre, it will be in neutral or indifferent
equilibrium.
Transverse Metacentre. We shall deal first with
transverse inclinations, because they are the more important,
and deal with inclinations in a longitudinal or foreandaft
direction in the next chapter.
Let Fig. 45 represent the section of a ship steadily inclined
STABLE.
Fir,. 45.
at a small angle from the upright by some external force,
such as the wind. The vessel has the same weight before and
after the inclination, and consequently has the same volume
of displacement. We must assume that no weights on board
shift, and consequently the centre of gravity remains in the
same position in the ship. But although the total volume of
Conditions of Equilibrium, Transverse Metacentre, etc. 95
displacement remains the same, the shape of this volume
changes, and consequently the centre of buoyancy will shift
from its original position. In the figure the ship is repre
sented by the section, WAL being the immersed section
when upright, WL being the position of the waterline on
the ship. On being inclined, WL' becomes the waterline,
and WAL' represents tbe immersed volume of the ship, which,
although different in shape, must have the same volume as the
original immersed volume WAL.
The wedgeshaped volume represented by WSW, which
has come out of the water, is termed the " emerged" or " out"
wedge. The wedgeshaped volume represented by LSL',
which has gone into the water, is termed the "immersed" or
" in " wedge. Since the ship retains the same volume of
displacement, it follows that the volume of the emerged wedge
WSW is equal to the volume of the immersed wedge LSL'.
It is only for small angles of inclination that the point S,
where the waterlines intersect, falls on the middle line of the
vessel. For larger angles it moves further out, as shown in
Fig. 77.
Now consider the vessel inclined at a small angle from
the upright, as in Fig. 45. The new volume of displacement
WAL' has its centre of buoyancy in a certain position, say B'.
This position might be calculated from the drawings in the
same manner as we found the point B, the original centre of
buoyancy ; but we shall see shortly how to fix the position of
the point B' much more easily.
B' being the new centre of buoyancy, the upward force of
the buoyancy must act through B', while the weight of the ship
acts vertically down through G, the centre of gravity of the
ship. Suppose the vertical through B' cuts the middle line of
the ship in M ; then we shall have two equal forces acting on
the ship, viz.
(1) Weight acting vertically down through the centre of
gravity.
(2) Buoyancy acting vertically up through the new centre
of buoyancy.
But they do not act in the same vertical line. Such a system
9 6
Theoretical Naval Architecture.
of forces is termed a couple. Draw GZ perpendicular to the
vertical through B'. Then the equal forces act at a distance
from each other of GZ. This distance is termed the arm of
the couple, and the moment of the couple is W X.GZ. On
looking at the figure, it is seen that the couple is tending to
take the ship back to the upright If the relative positions
of G and M were such that the couple acted as in Fig. 46,
the couple would tend to take the ship farther away from the
upright ; and again, if G and M coincided, we should have the
forces acting in the same vertical line, and consequently no
UNSTABLE. /
FIG. 46.
couple at all, and the ship would have no tendency to move
either to the upright or away from it.
We see, therefore, that for a ship to be in stable equilibrium
for any direction of inclination, it is necessary that the point
M be above the centre of gravity of the ship. This point M
is termed the metacentre. We now group together the three
conditions which must be fulfilled in order that a ship may
float freely and at rest in stable equilibrium
(i) The weight of water displaced must equal the total
weight of the ship (see p. 23).
Conditions of Equilibrium, Transverse Metacentre, etc. 97
(2) The centre of gravity of the ship must be in the same
vertical line as the centre of gravity of the displaced water
(centre of buoyancy) (see p. 93).
(3) The centre of gravity of the ship must be below the
metacentre.
For small transverse inclinations, M is termed the transverse
metacentre^ which we may accordingly define as follows :
For a given plane of flotation of a vessel in the upright
condition, let B be the centre of buoyancy, and BM the vertical
through it. Suppose the vessel inclined transversely through
a very small angle, retaining the same volume of displacement,
B' being the new centre of buoyancy, and B'M the vertical
through it, meeting BM in M. Then this point of intersection,
M, is termed the transverse metacentre.
There are two things in this definition that should be noted :
(i) the angle of inclination is supposed very small, and (2) the
volume of displacement remains the same.
It is found that, for all practical purposes, in ordinary ships
the point M does not change in position for inclinations up to
as large as 10 to 15; but beyond this it takes up different
positions.
We may now say, with reference to a ship's initial stability
or stability in the upright condition
(1) If G is below M, the ship is in stable equilibrium.
(2) If G is above M, the ship is in unstable equilibrium.
(3) If G coincides with M, the ship is in neutral or in
different equilibrium.
We thus see how important the relative positions of the
centre of gravity and the transverse metacentre are as affecting
a ship's initial stability. The distance GM is termed the
transverse metacentric height^ or, more generally, simply the
metacentric height.
We have seen that for small angles M remains practically
in a constant position, and consequently we may say GZ
= GM . sin 6 for angles up to 10 to 15, say. GZ is the
arm of the couple, and so we can say that the moment of the
couple is
W x GM . sin B
H
9$ Theoretical Naval Architecture.
If M is above G, this moment tends to right the ship, and
we may therefore say that the moment of statical stability at the
angle is
W X GM . sin
This is termed the metacentric method of determining a
vessel's stability. It can only be used at small angles of
inclination to the upright, viz. up to from 10 to 15 degrees.
Example. A vessel of 14,000 tons displacement has a metacentric
height of 3^ feet. Then, if she is steadily inclined at an angle of 10, the
tendency she has to return to the upright, or, as we have termed it, the
moment of statical stability, is
14,000 x 3 '5 X sin 10 = 8506 foottons
We shall discuss later how the distance between G and M,
or the metacentric height, influences the behaviour of a ship,
and what its value should be in various cases ; we must now
investigate the methods which are employed by naval archi
tects to determine the distance for any given ship.
There are two things to be found, viz. (i) the position of
G, the centre of gravity of the vessel ; (2) the position of M,
the transverse metacentre.
Now, G depends solely upon the vertical distribution of the
weights forming the structure and lading of the ship, and the
methods employed to find its position we shall deal with
separately ; but M depends solely upon the form of the ship,
and its position can be determined when the geometrical form
of the underwater portion of the ship is known. Before we
proceed with the investigation of the rules necessary to do this,
we must consider certain geometrical principles which have to.
be employed.
Centre of Flotation. If a floating body is slightly
inclined so as to maintain the same volume of displacement,
the new waterplane must pass through the centre of gravity of
the original waterplane. In order that the same volume of
displacement may be retained, the volume of the immersed
wedge SLL 1} Fig 47, must equal the volume of the emerged
wedge SWWj. Call y an ordinate on the immersed side, and
y' an ordinate on the emerged side of the waterplane. Then
Conditions of Equilibrium, Transverse Metacentre, etc. 99
the areas of the sections of the immersed and emerged wedges
are respectively (since LI^ = y . dO, WWi = / . dO t dQ being
the small angle of inclination) .
*/.<#, ' *(/)' <#
and using the notation we have already employed
Volume of immersed wedge = %fy z . dO .dx
emerged =Wf.M.dx
ind accordingly
.dd.dx
SECTION. 
or
But J/T* . dx is the moment of the immersed portion of the
waterplane about the intersection,
and i/(y) 2 . dx is the moment of
the emerged portion of the water
plane about the intersection (see
p. 59) ; therefore the moment of
one side of the waterplane about
the intersection is the same as the
moment of the other side, and
consequently the line of inter
section passes through the centre
of gravity of the waterplane.
The centre of gravity of the water
plane is termed the centre of flota
tion. In whatever direction a
ship is inclined, transversely,
longitudinally, or in any interme
diate direction, through a small
angle, the line of intersection of
the new waterplane with the
original waterplane must always
pass through the centre of flotation. For transverse inclinations
of a ship the line of intersection is the centre line of the water
plane ; for longitudinal inclinations the foreandaft position of
the centre of flotation has to be calculated, as we shall see
when we deal with longitudinal inclinations.
TOO
Theoretical Naval Architecture.
w
Shift of the Centre of Gravity of a Figure due to
the Shift of a Portion of the Figure. In Fig. 48 the
figure PQRSTU is made up of the two portions PQTU and
QRST, with centres of
S gravity at g and g / respec
tively. Let a, d be the
areas, the whole area a f a'
= A. Then the C.G. of
the whole area is at G, such
that a X gG = a' x /G, or
g=J U the C.G. di
vides the line joining g and
g' inversely as the areas.
If now the portion QRST
is shifted to the position
UTVW with C.G. g\ the
FIG 48.
C.G. of the new combination PQWV is on the line^g" at G'
such that
^G' = a = /G*
Therefore by the properties of triangles GG' is parallel to g'g".
Also
Now, taking moments about g we have a X gg 1 = A x ^G
. GG' = a
"g'g" A
or GG', the shift of the C.G. = ^ x g'g"
or the whole area multiplied by its shift equals the small area
multiplied by its shift, and these shifts are in parallel directions.
Also for the horizontal shift, A X gG' = a x gg", and for the
vertical shift, A X G^ = a X g'g. The above proof is perfectly
general, although a simple figure has been taken by which its
truth may be readily seen. It applies equally to the shift of
weights.
Conditions of Equilibrium, Transverse Metacentre, etc. IOI
The uses that are made of this will become more apparent
as we proceed, but the following examples will serve as illus
trations :
Example. A vessel weighing W tons has a weight w tons on the deck.
This is shifted transversely across the deck a distance of d feet, as in Fig. 49.
Find the shift of the C.G. of the vessel both in direction and amount.
W
A
FIG. 49.
G will move to G' such that GG' will be parallel to the line joining
the original and final positions of the weight w ;
If w = 70 tons, d = 30 feet, W = 5000 tons, then
GG' = 7 X3 = f& feet = 042 foot
5000
Example. In a vessel of 4000 tons displacement, suppose 100 tons ol
coal to be shifted so that its C.G. moves 18 feet transversely and 4^ feet
vertically. Find the shift of the C.G. of the vessel.
The C.G. will move horizontally an amount equal to  = 0*45 ft.
and vertically an amount equal to  = O' 1 1 ft.
Moment of Inertia. We have dealt in Chapter II. with
102
Theoretical Naval Architecture.
the moment of a force about a given point, and we denned it as
the product of the force and the perpendicular distance of its
line of action from the point ; also the moment of an area
about a given axis as being the area multiplied by the distance
of its centre of gravity from the axis. We could find the
moment of a large area about a given axis by dividing it into
a number of small areas and summing up the moments of all
these small areas about the axis. In this we notice that the
area or force is multiplied simply by the distance. Now we
have to go a step further, and imagine that each small area is
multiplied by the square of its distance from a given axis. If
all such products are added together for an area, we should
obtain not the simple moment, but what may be termed the
01
y
FIG. 50.
moment of the second degree, or more often the moment of
inertia of the area about the given axis. 1 We therefore define
the moment of inertia of an area about a given axis as
follows :
1 This is the geometrical moment of inertia. Strictly speaking, moment
of inertia involves the mass of the body. We make here the same assump
tion that we did in simple moments (p. 49), viz. that the area is the
surface of a very thin lamina or plate of homogeneous material of uniform
thickness.
Conditions of Equilibrium, Transverse Metacentre, etc. 103
Imagine the. area divided into very small areas, and each such
small area multiplied by the square of its distance from the given
axis ; then, if all these products be added together, we shall obtain
the moment of inertia of the total area about the given axis.
Thus in Fig. 50, let OO be the axis. Take a very small
area, calling it //A, distance^ from the axis. Then the sum
of all such products as dh. x 7 2 , or (using the notation we have
employed) // 2 . dA, will be the moment of inertia of the area
about the axis OO.
To determine this for any figure requires the application of
advanced mathematics, but the result for certain regular figures
are given below.
It is found that we can always express the moment of
inertia, often written I, of a plane area about a given axis by
the expression
where A is the area of the figure ;
h is the depth of the figure perpendicular to the axis ;
n is a coefficient depending on the shape of the figure
and the position of the axis.
First, when the axis is through the centre of gravity of
the figure parallel to the base, as in Figs. 51 and 52
**.
l
FIG 51.
for a circle n
for a rectangle n
for a triangle
FIG. 52.
r, so that I = yV
r, I =
I
1 9 B * "" "
IO4 Theoretical Naval Architecture.
Second, when the axis is one of the sides
for a rectangle n = J, so that I =
for a triangle n = , I =
Example. Two squares of side a are joined to form a rectangle. The
I of each square about the common side is
J(a 2 )a 2 (a 1 area)
the I of both about the common side will be the sum of each taken
separately, or
If, however, we took the whole figure and treated it as a rectangle, its I
about the common side would be
^(2fl')(2a) = \a> (area = 2a*)
which is the same result as was obtained before.
To find the moment of inertia of a plane figure about an axis
parallel to and a given distance from an axis through its centre
of gravity.
Suppose the moment of inertia about the axis NN passing
through the centre of gravity of the figure (Fig. 53) is I , the
0.
lit
J
fa
J j
JN.
0.
area of the figure is A, and
OO, the given axis, is parallel
to NN and a distance y
from it. Then the moment
of inertia (I) of the figure
about OO is given by
I = I + A/
The moment of inertia of an
area about any axis is there
fore determined by adding
to the moment of inertia of
the area about a parallel axis
through the centre of gravity,
the product of the area into
the square of the distance
between the two axes. We
FIG. S3
see from this that the moment of inertia of a figure about an
axis through its own centre of gravity is always less than about
any other axis parallel to it.
Conditions of Equilibrium, Transverse Metacentre etc. 105
Example. Having given the moment of inertia of the triangle in
Fig. 52 about the axis NN through the centre of gravity as T gA/fc 2 , find the
moment of inertia about the base parallel to NN.
Applying the above rule, we have
A
which agrees with the value given above for the moment of inertia of a
triangle about its base.
Example. Find the moment of inertia of a triangle of area A and
height h about an axis through the vertex parallel to the base.
Ans. JA^ 2 .
Example. A rectangle is 4 inches long and 3 inches broad. Compare
the ratio of its moment of inertia about an axis through the centre parallel
to the long and short sides respectively.
Ans. 9 : 16.
Example. A square of 12 inches side has another symmetrical square
of half its area cut out of the centre. Compare the moments of inertia
about an axis through the centre parallel to one side of, the original
square, the square cut out, the remaining area.
Ans. As 4 : i I 3, the ratio of the areas being 4:252.
This last example illustrates the important fact that if an
area is distributed away from the centre of gravity, the moment
of inertia is very much greater than if the same area were
massed near the centre of gravity.
To find the Moment of Inertia of a Plane Cur
vilinear Figure (as Fig. 36, p. 59) about its Base. Take
a strip PQ of length y and breadth (indefinitely small) dx>
Then, if we regard PQ as a rectangle, its moment of inertia
about the base DC is
\(y . dx)y^ y 3 . dx ( y . dx = area)
and the moment of inertia of the whole figure about DC will
be the sum of all such expressions as this ; or
that is, we put the third part of the cubes of the ordinates of the
curve through either of Simpson's rules. For the waterplane
of a ship (for which we usually require to find the moment of
inertia about the centre line), we must add the moment of
inertia of both sides together: and, since these are symmetrical,
we have
1 = I /y d* (y = semiordinate of waterplane).
io6
Theoretical Naval Architecture
In finding the moment of inertia of a waterplane about the
centre line, the work is arranged as follows :
Number of
ordinate.
Semiordinates
of waterplane.
Cubes of
semiordinates.
Simpson's
multipliers.
Functions of
cubes.
T
005
,
2
4'65
IOI
4
404
3
1005
1015
2
2,030
4
I4'3O
2924
4.
11,696
5
16*75
4699
2
9,398
6
1765
5498
4
21,992
7
1740
5268
2
10,536
8
l6'2O
4252
4
17,003
9
1355
2488
2
4,976
10
899
4
3,596
ii
365
49
I
49
81,685
Common interval = 28 feet
Moment of inertia = 81,685 X f X ^ = 508,262 x
The semiordinates are placed in column 2, and the cubes
of these are placed in column 3. It is not necessary, in ordi
nary cases, to put any decimal places in the cube ; the nearest
whole number is sufficient. A table of cubes is given in
the Appendix for numbers up to 45, rising by 0*05. These
cubes are put through Simpson's multipliers in the ordinary
way, giving column 5. The sum of the functions of cubes
has to be treated as follows : First there is the multiplier for
Simpson's rule, viz. \ x 28, and then the of the expres
sion /_y 3 . dx> which takes into account both sides. The
multiplier, therefore, is f X ", and the sum of the numbers
in column 5 multiplied by this will give the moment of inertia
required.
Approximation to the Moment of Inertia of a
Ship's Waterplane about the Centre Line. We have
seen that for certain regular figures we can express the moment
of inertia about an axis through the centre of gravity in the
form A/fc 2 , where n is a coefficient varying for each figure.
We can, in the same way, express the I of a waterplane area
1 This calculation for the L.W.P. is usually done on the displacement
sheet. Brown's displacement sheet provides a column for each waterliue
except the two lowest.
Conditions of Equilibrium . Transverse Metacentre, etc. \ 07
about the centre line, but it is not convenient to use the area
as we have done above. We know that the area can be
expressed in the form
k x L X B
where L is the extreme length ;
B breadth;
k is a coefficient of fineness ;
so that we can write
I = LB 8
where n is a new coefficient that will vary for different shapes
of waterplanes. If we can find what the values of the co
efficient n are for ordinary waterplanes, it would be very
useful in checking our calculation work. Taking the case of
a L.W.P. in the form of a rectangle, we should find that n =
0*08, and for a L.W.P. in the form of two triangles, n = 0*02.
These are two extreme cases, and we should expect for
ordinary ships the value of the coefficient n would lie between
these values. This is found to be the case, and we may take
the following approximate values for the value of n in the
formula I = nLE 3 :
For ships whose load waterplanes are extremely fine ... 0*04
,, ,, ,, moderately tine ... 0*05
,, ,, ,, ,, very full ... ... O'o6
For the waterplane whose moment of inertia we calcu
lated above, we have, length 280 feet, breadth 353 feet, and
I = 508,262 in footunits. Therefore the value of the coefficient
n is
508262
280 x (353)' = ' 41
Formula for finding the Distance of the Trans
verse Metacentre above the Centre of Buoyancy
(BM). We have already discussed in Chapter II. how the
position of the centre of buoyancy can be determined if the under
water form of the ship is known, and now we proceed to discuss
how the distance BM is found. Knowing this, we are able to
fix the position of the transverse metacentre in the ship.
io8 Theoretical Naval Architecture.
Let Fig. 45, p. 94, represent a ship heeled over to a very
small angle 6 (much exaggerated in the figure).
B is the centre of buoyancy in the upright position when
floating at the waterline WL.
B' is the centre of buoyancy in the inclined position when
floating at the waterline W'L'.
v is the volume of either the immersed wedge LSL or the
emerged wedge WSW.
V is the total volume of displacement.
g is the centre of gravity of the emerged wedge.
g is the centre of gravity of the immersed wedge.
Then, using the principle given on p. 100, BB' will be parallel
to gg, and
since the new displacement is formed by taking away the wedge
WSW from the original displacement, and putting it in the
position LSL'.
Now for the very small angle of inclination, we may say
tliat
BB'
BM = sln *
or BB' = BM sin
so that we can find BM if we can determine the value of
v x gg t since V, the volume of displacement, is known.
Let Fig. 54 be a section of the vessel; //, //', the original
and new waterlines respectively, the angle of inclination being
very small. Then we may term wSu/ the emerged triangle,
and /S/ the immersed triangle, being transverse sections of
the emerged and immersed wedges, and a/a/, // being for all
practical purposes straight lines. If y be the halfbreadth of
the waterline at this section, we can say ww' = II' = y sin 0,
and the area of either of the triangles is
\y X y sin = ^y z sin 6
Let 0, d be the centres of gravity of the triangles #>Sze/, /S/
respectively ; then we can say, seeing that 6 is very small, that
Conditions of Equilibrium, Transverse Metacentre, etc. 109
ad = f.y, since the centre of gravity of a triangle is twothirds
the height from the apex. The new immersed section being
regarded as formed by the transference of the triangle
FIG.
to the position occupied by the triangle /S/, the moment of
transference is
(l/ sin 0) X & = f / sin
and for a very small length dx of the waterline the moment
will be
f y sin . dx
since the small volume is \f sin . dx, and the shift of its
centre of gravity is f jy. If now we summed all such expres
sions as this for the whole length of the ship, we should get
the moment of the transference of the wedge, or v x gg'.
Therefore we may say, using the ordinary notation
vxgg' = /fy sin e . dx
= fsin0jy.dk
therefore we have
v X gg __  sin jy . dx
~V ~~ " v
BB' = BM sin =
or BM =
But the numerator of this expression is what we have found to
no Theoretical Naval Architecture.
be the moment of inertia of 'a waterplane about its centre line,
y being a semiordinate ; therefore we can write
We have seen, on p. 105, how the moment of inertia of a
waterplane is found for any given case, and knowing the
volume of displacement, we can then determine the distance
BM, and so, knowing the position of the C.B., fix the position
of the transverse metacentre in the ship.
Example. A lighter is in the form of a box, 120 feet long, 30 feet
broad, and floats at a draught of 10 feet. Find its transverse BM.
In this case the waterplane is a rectangle 120' X 30', and we want its
I about the middle line. Using the formula for the I of a rectangle about
an axis through its centre parallel to a side, nA/4 2 , we have
1 = n X 3 6 oo X 900 (k = 30)
= 270,000
V, the volume of displacement, = 120x30x10 = 36,000
BM = 270,000 5feet
36,000
Example. A pontoon of 10 feet draught has a constant sectio*n in the
form of a trapezoid, breadth at the waterline 30 feet, breadth at base
20 feet, length i'2O feet. Find the transverse BM.
Ans. 9 feet.
It will be noticed that the waterplane in this question is
the same as in the previous question, but the displacement being
less, the BM is greater. M is therefore higher in the ship for
two reasons. BM is greater and B is higher in the second case.
Example. A raft is formed of two cylinders 5 feet in diameter, parallel
throughout their lengths, and 10 feet apart, centre to centre. The raft floats
with the axes of the cylinders in the surface. Find the transverse BM.
We shall find that the length does not affect the result, but we will
suppose the length is / feet. We may find the I of the waterplane in two
ways. It consists of two rectangles each /' X 5', and their centre lines
are 10 feet apart.
1. The water plane may be regarded as formed by cutting a rectangle
/' x 5' out of a rectangle /' X 15' ;
.. I  (/ x 15) x is 2  A(/ x 5) x s 3
= Mi5 8  5 3 )
= afV
this being about a foreandaft axis at the centre of the raft.
2. We may take the two rectangles separately, and find the I of each
about the centre line of the raft, which is 5 feet from the line through the
centre of each rectangle. Using the formula
I = I. + Ay'
(/ x 5)5* + (/ x 5)5*
Conditions of Equilibrium^ Transverse Metacentre, etc. 1 1 1
and for both rectangles the moment of inertia will be twice this, or 3 f 8 /, as
obtained above.
We have to find the volume of displacement, which works out to *$l
cubic feet. The distance BM is therefore
s$o/4^./= 1 3 8 feet
Example. A raft is formed of three cylinders, 5 feet in diameter,
parallel and symmetrical throughout their lengths, the breadth extreme
being 25 feet. The raft floats with the axes of the cylinders in the surface.
Find the transverse BM.
The moment of inertia of the waterplane of this raft is best found by
using the formula I = I + Ay 8 for the two outside rectangles, and adding
it to I , the moment of inertia of the centre rectangle about the middle line.
We therefore have for the whole waterplane I = <\p/, where / = the
length ; and the volume of displacement being  8 2 2 8 5 /, the value of BM will be
35 ^et.
Approximate Formula for the Height of the Trans
verse Metacentre above the Centre of Buoyancy.
The formula for BM is
We have seen that we may express I as LB 8 , where n is
a coefficient which varies for different shapes of waterplanes,
but which will be the same for two ships whose waterplanes
are similar.
We have also seen that we may express V as LBD, where
D is the mean moulded draft (to top of keel amidships), and k
is a coefficient which varies for different forms, but which will
be the same for two ships whose under water forms are similar.
Therefore we may say
n X L X B 3
BM =
B 2
where a is a coefficient obtained from the coefficients n and k.
Sir William White, in the "Manual of Naval Architecture,"
gives the value of a as being between 0*08 and 0*1, a usual
value for merchant ships being 0*09. The above formula
shows very clearly that the breadth is more effective than the
draught in determining what the value of BM is in any given
case. It will also be noticed that the length is not brought in.
112
Theoretical Naval Architecture,
The ship for which the moment of inertia of a waterplane
was calculated on p. 106, had a displacement of 1837 tons up
to that waterplane. The value of BM is therefore
508262
X 35
7*91 feet
The breadth and mean draught were 35*3 and 13! feet re
spectively. Consequently the value of the coefficient a is
0*084.
To prove that a Homogeneous Log of Timber of
Square Section and Specific Gravity P 5 cannot float
in Fresh Water with One of its Faces Horizontal.
The log having a specific gravity of 05 will float, and will float
with half its substance immersed. The condition that it shall
float in stable equilibrium, as regards transverse inclination, in
any position is that the transverse metacentre shall be above
the centre of gravity.
Let the section be as indicated in Fig. 55, with side length
20. And suppose the log
is placed in the water with
one side of this section
horizontal. Then the
draughtline will be at a
distance a from the bot
tom, and the log, being
homogeneous, i.e. of the
same quality all through,
will have its C.G. in the
middle at G, at a distance
also of a from the bottom.
The centre of buoyancy
FIG. 55. will be at a distance of
 from the bottom. The height of the transverse metacentre
2
above the centre of buoyancy is given by
! I
j 42a,
BM
I
V
Conditions of Equilibrium, Transverse Metacentre, etc. 113
where I = moment of inertia of waterplane about a longi
tudinal axis through its centre
V = volume of displacement in cubic feet.
Now, the waterplane of the log is a rectangle of length /
and breadth 20, and therefore
its I = j /. za(2of
and V = /. 2a . a = zla?
:. BM = iV^ 3 r 2la* = a
But BG = \a
therefore the transverse metacentre is below the centre of
gravity, and consequently the log cannot float in the position
given.
If, now, the log be assumed floating with one corner down
ward, it will be found by a precisely similar method that
BG = 04710
and BM = 0*9430
Thus in this case the transverse metacentre is above the
centre of gravity, and consequently the log will float in stable
equilibrium.
It can also be shown by similar methods that the position
of stable equilibrium for all directions of inclination of a cube
composed of homogeneous material of specific gravity o'5 is
with one corner downwards.
Metacentric Diagram. We have seen how the position
of the transverse metacentre can be determined for any given
ship floating at a definite waterline. It is often necessary,
however, to know the position of the metacentre when the ship
is floating at some different waterline ; as, for instance, when
coal or stores have been consumed, or when the ship is in a light
condition. It is usual to construct a diagram which will show
at once, for any given mean draught which the vessel may have,
the position of the transverse metacentre. Such a diagram is
shown in Fig. 56, and it is constructed in the following manner:
A line W^ is drawn to represent the load waterline, and
parallel to it are drawn W 2 L 2 , W 3 L 2 , W 4 L 4 to represent the
i
Theoretical Naval Architecture.
waterlines Nos. 2, 3, and 4, which are used for calculating the
displacement, the proper distance apart, a convenient scale
being \ inch to i foot. A line L^ 4 is drawn cutting these
level lines, and inclined to them at an angle of 45. Through
the points of intersection L 1} L 2 , L 3 , L 4 , are drawn vertical lines
as shown. The ship is then supposed to float successively at
these waterlines, and the position of the centre of buoyancy
and the distance of the transverse metacentre above the C.B.
il
W,
w
FIG. 56.
calculated for each case. The methods employed for finding
the position of the C.B. at the different waterlines have already
been dealt with in Chapter II. On the vertical lines are then
set down from the L.W.L. the respective distances of the
centres of buoyancy below the L.W.L. Thus L^ is the
distance when floating at the L.W.L., and AB 3 the distance
when floating at No. 3 W.L. In this way the points B 15 B 2 ,
B 3 , B 4 are obtained ; and if the calculations are correct, a fair
Conditions of Equilibrium, Transverse Metacentre, etc. 115
line can be drawn passing through all these spots as shown.
Such a curve is termed the curve of centres of buoyancy. It is
usually found to be rather a flat curve, being straight near the
loadline condition. The distance BM for each waterline is
then set up from Bi, B 2 , B 3 , B 4 respectively, giving the points
MU M 2 , M 3 , M 4 . A curve can then be drawn through these
points, which is termed the curve of transverse metacentres.
Now, suppose the ship is floating at some intermediate water
line say wl: through /, where wl cuts the 45 line, draw a
vertical cutting the curves of centres of buoyancy and meta
centres in ]b and m respectively. Then m will be the position
of the transverse metacentre of the ship when floating at the
waterline wl. It will be noticed that we have supposed the
ship to float always with the waterplane parallel to the L.W.P.;
that is to say, she does not alter trim. For waterplanes not
parallel to the L.W.P. we take the mean draught (i.e. the
draughts at the foreandaft perpendiculars are added together
and divided by 2), and find the position of M on the meta
centric diagram for the waterplane, parallel to the L.W.P.,
corresponding to this mean draught. Unless the change of
trim is very considerable, this is found to be correct enough
for all practical purposes. Suppose, however, the ship trims
very much by the stern, 1 owing to coal or stores forward being
consumed, the shape of her waterplane will be very different
from the shape it would have if she were floating at her normal
trim or parallel to the L.W.P. ; generally the waterplane will
be fuller under these circumstances, and the moment of inertia
will be greater, and consequently M higher in the ship, than
would be given on the metacentric diagram. When a ship
is inclined, an operation that will be described later, she
is frequently in an unfinished condition, and trims consider
ably by the stern. It is necessary to know the position of
the transverse metacentre accurately for this condition, and
1 This would be the case in the following : A ship is designed to float
at a draught of 17 feet forward and 19 feet aft, or, as we say, 2 feet by the
stern. If her draught is, say, 16 feet forward and 20 feet aft, she will have
the same mean draught as designed, vk. 18 feet, but she will trim 2 feet
more by the stern.
1 1 6 Theoretical Naval A rchitecture.
consequently the metacentric diagram cannot be used, but a
separate calculation made for the waterplane at which the
vessel is floating.
On the metacentric diagram is placed also the position of
the centre of gravity of the ship under certain conditions. For
a merchant ship these conditions may vary considerably owing
to the nature of the cargo carried. There are two conditions
for which the C.G. may be readily determined, viz. the light
condition, and the condition when loaded to the loadline with
a homogeneous cargo. The light condition may be denned as
follows : No cargo, coal, stores, or any weights on board not
actually forming a part of the hull and machinery, but includ
ing the water in boilers and condensers. The draughtlines for
the various conditions are put on the metacentric diagram, and
the position of the centre of gravity for each condition placed
in its proper vertical position. The various values for GM, the
metacentric height, are thus obtained.
On the left of the diagram are placed, at the various water
lines, the mean draught, displacement, and tons per inch. 1
There are two forms of section for which it is instructive to
construct the metacentric diagram.
1. A floating body of constant rectangular section.
2. A floating body of constant triangular section, the apex
of the triangle being at the bottom.
i. For a body having a constant rectangular section, the
moment of inertia of the waterplane is the same for all
draughts, but the volume of displacement varies. Suppose the
rectangular box is 80 feet long, 8 feet broad, 9 feet deep. Then
the moment of inertia of the waterplane for all draughts is
5*5(80 X 8) x 8" = ^4P
The volumes of displacement are as follows :
Draught 6 inches V = 80 X 8 X J cubic feet
1 foot V = 8ox8
2 feet V = 80x8x2
4 V = Sox 8x4 ..
V = 80x8x7
" t 9 V =8ox 8x9
1 For a specimen metacentric diagram, see Example 40, Chap. III.
Specimen diagrams for various types of ships are given in the Author's
"War Ships."
Conditions of Equilibrium, Transverse Metacentre, etc. 1 1 7
and the values of BM are therefore as follows :
Draught 6 inches BM = 10*66 feet
1 foot ... ' BM = 533
2 feet BM = 2'66
4 BM= 133
7 ,, BM = 076
9 BM = 059
The centre of buoyancy is always at halfdraught, so that
its locus or path will be a straight line, 1 and if the values obtained
above are set off from the centres of buoyancy at the various
waterlines, we shall obtain the curve of transverse metacentres
as shown in Fig. 57 by the curve A A, the line BB being the
corresponding locus of the centres of buoyancy.
90.
6O.
35.
Ox'
FIG. 57
s. For a floating body with a constant triangular section, the
locus of centres of buoyancy is also a straight line because it is
always twothirds the draught above the base. 1 Suppose the
triangular section to be 10 feet broad at the top and 9 feet deep,
the length of the body being 120 feet. In this case we must
calculate the moment of inertia of each waterplane and the
volume of displacement up to each. The results are found to
be as follows :
' This may be seen by finding a few spots on this locus.
uS Theoretical Naval Architecture.
Draught I foot BM = 0*20 feet
,, 2 feet BM = 0*41 ,,
4 BM = o82
6 BM=i2 3
9 BM = i85
These values are set up from the respective centres of
buoyancy, and give the locus of transverse metacentres, which
is found to be a straight line, as shown by CC in Fig. 57, DD
being the locus of centres of buoyancy.
Approximation to Locus of Centres of Buoyancy
on the Metacentric Diagram. We have seen (p. 65) how
the distance of the centre of buoyancy below the L.W.L. can
be approximately determined. The locus of centres of buoyancy
in the metacentric diagram is, in most cases, very nearly straight
for the portion near the loadline, and if we could obtain easily
the direction the curve takes on leaving the position for the load
waterline, we should obtain a very close approximation to the
actual curve itself. It might be desirable to obtain such an
approximation in the early stages of a design, when it would
not be convenient to calculate the actual positions of the centre
of buoyancy, in order to accurately construct the curve.
Let be the angle the tangent to the curve of buoyancy at
the load condition makes with the horizontal, as in
Fig. 56;
A, the area of the load waterplane in square feet ;
V, the volume of displacement up to the load waterline
in cubic feet ;
^, the distance of the centre of buoyancy of the load
displacement below the load waterline in feet.
Then the direction of the tangent to the curve of buoyancy is
given by
tan 9 = * (for proof see later.)
Each of the terms in the latter expression are known or can
be readily approximated to, 1 and we can thus determine the in
clination at which the curve of centres of buoyancy will start,
and this will closely follow the actual curve. 2
1 See Example 39, p. 143, for a further approximation.
9 See a paper by the late Professor Jenkins read before the Institution
of Naval Architects in 1884.
Conditions of Equilibrium, Transverse Metacentre, etc. \ 19
In a given case
A = 7854 square feet
h = 5*45 f e et
V = 2140 x 35 cubic feet
so that
2140 x 35 i
= 0572*
Finding the Metacentric Height by Experiment.
Inclining Experiment. We have been dealing up to the
present with the purely geometrical aspect of initial stability,
viz. the methods employed and the principles involved in
finding the position of the transverse metacentre. All that is
needed in order to determine this point is the form of the
underwater portion of the vessel. But in order to know any
thing about the vessel's initial stability, we must also know the
vertical position of the centre of gravity of the ship, and it is to
determine this point that the inclining experiment is performed.
This is done as the vessel approaches completion, when
weights that have yet to go on board can be determined
together with their final positions. Weights are shifted trans
versely across the deck, and by using the principle explained on
p. 100, we can tell at once the horizontal shift of the centre of
gravity of the ship herself due to this shift of the weights on
board. The weight of the ship can be determined by calculating
the displacement up to the waterline she floats at, during the
experiment. (An approximate method of determining this
displacement when the vessel floats out of her designed trim
1 The best way to set off this line is to set off a horizontal line of 10 feet
long (on a convenient scale), and from the end set down a vertical line
572 feet long on the same scale. This will give the inclination required,
for tan = f^ = ^ = 0572.
base 10
This remark applies to any case in which an angle has to be set off
very accurately. A table of tangents is consulted and the tangent of
the required angle is found, and a similar process to the above is gone
through.
I2O
Theoretical Naval Architecture.
will be found on p. 152.) Using the notation employed on
p. 100, and illustrated by Fig. 49, we have
rr , wxd
GG = IT"
Now, unless prevented by external forces, it is evident that
the vessel must incline over to such an angle that the centre of
gravity G' and the centre of buoyancy B' are in the same verti
cal line (see Fig. 58), and, the angle of inclination being small,
FIG. 58.
M will be the transverse metacentre. If now we call 6 the
angle of inclination to the upright, GM being the " metacentric
height "
GG'
w X d
~ W X tan
using the value found above for GG'. The only term that we
do not yet know in this expression is tan 6, and this is found in
the following manner : At two or three convenient positions
Conditions of Equilibrium, Transverse Metacentre, etc. 1 2 1
in the ship x (such as at bulkheads or down hatchways) plumb
bobs are suspended from a point in the middle line of the ship,
and at a convenient distance from the point of suspension a
horizontal batten is fixed, with the centre line of the ship marked
on it, as shown by PQ in Fig. 58. Before the ship is inclined,
the plumbline should coincide, as nearly as possible, with the
centreline of the ship that is to say, the ship should be prac
tically upright. When the ship is heeled over to the angle 0,
the plumbline will also be inclined at the same angle, 0, to the
original vertical or centre line of the ship, and if / be the
distance of the horizontal batten below the point of suspension
O in inches, and a the deviation of the plumbline along the
batten, also in inches, the angle 6 is at once determined, for
tan 6 = ,
so that we can write l
In practice it is convenient to check the results obtained, by
dividing the weight w into four equal parts, placing two sets on
one side and two sets on the other side, arranged as in Fig. 59.
The experiment is then performed in the following order :
(a) See if the ship is floating upright, in which case the
plumblines will coincide with the centre of the ship.
(b) The weight (i), Fig. 59, is shifted from port to star
board on to the top of weight (3) through the distance d feet,
say, and the deviations of the plumblines are noted when the
ship settles down at a steady angle.
(c) The weight (2) is shifted from port to starboard on to
the top of weight (4) through the distance d feet, and the
deviations of the plumbline noted.
(d) The weights (i) and (2) are replaced in their original posi
tions, when the vessel should again resume her upright position.
* If two positions are taken, one is forward and the other aft. If three
positions are taken, one is forward, one aft, and one amidships.
* This depends on the assumption that M is a fixed point for the heel
obtained, and this is true for ordinary ships. It fails, however, in the case
of a vessel of very small or of zero metacentric height. See examples in
Appendix A.
122 Theoretical Naval Architecture.
(e) The weight (3) is moved from starboard to port, and
the deviations of the plumblines noted.
(/) The weight (4) is moved from starboard to port, and
the deviations of the plumblines noted.
With the above method of conducting the experiment, 1 and
using two plumblines, we obtain eight readings, and if three
plumblines were used we should obtain twelve readings. It is
important that such checks should be obtained, as a single result
might be rendered quite incorrect, owing to the influence of the
hawsers, etc. A specimen experiment is given on p. 123, in
which two plumblines were used. The deviations obtained
!
a. j
_r* ~H
mm
mnh
i
1
1
!
FIG. 59.
are set out in detail, the mean deviation for a shift of 12^ tons
through 36 feet being 5 inches, or the mean deviation for a
shift of 25 tons through 36 feet is 10^ inches.
Precautions to be taken when performing an Inclining Experi
ment. A rough estimate should be made of the GM expected
at the time of the experiment \ the weight of ballast can then be
determined which will give an inclination of about 4 or 5 when
onehalf is moved a known distance across the deck. The weight
of ballast thus found can then be got ready for the experiment.
A personal inspection should be made to see that all weights
likely to shift are efficiently secured, the ship cleared of all
1 There is a slight rise of G, the centre of gravity of the ship, in this
method : but the error involved is inappreciable.
Conditions of Equilibrium, Transverse Metacentre, etc. 123
free water, and boilers either emptied or run up quite full.
Any floating stages should be released or secured by veiy slack
painters.
If possible a fine day should be chosen, with the water calm
and little wind. All men not actually employed on the experi
ment should be sent ashore. Saturday afternoon or a dinner
hour is found a convenient time, since then the majority of
the workmen employed finishing the ship are likely to be away.
The ship should be hauled head or stern on to the wind,
if any, and secured by hawsers at the bow and stern. When
taking the readings, these hawsers should be slacked out, so as
to ensure that they do not influence the reading. The ship
should be plumbed upright before commencing.
An account should be taken, with positions of all weights to
be placed on board to complete, of all weights to be removed,
such as yard plant, etc., and all weights that have to be shifted.
The following is a specimen report of an inclining experi
ment :
Report on Inclining Experiment performed on "
at . Density of water cubic feet to the ton.
, 189,
Draught of water
i j
Displacement in tons at this draught
1 6' 9" forward.
22' 10" aft.
5372
The wind was slight, and the ship was kept head to wind during the
experiment. Ballast used for inclining, 50 tons. Lengths of pendulums,
two in number, 15 feet. Shift of ballast across deck, 36 feet.
Deviation of pendulum in 15 feet.
Forward.
Aft.
Experiment I, 12^ tons port to starboard
>i 2, I2j ,, ,,
Ballast replaced, zero checked ...
Experiment 3, 12^ tons starboard to port
4, I2i
5g
10*"
right
$
5*"
io"
right
s;
I0j"
The condition of the ship at the time of inclining is as defined below :
Bilges dry.
Water tanks empty.
124 Theoretical Naval Architecture.
No water in boilers, feedtanks, condensers, distillers, cisterns, etc.
Workmen on board, 66.
Tools on board, 5 tons.
Masts and spars complete.
No boats on board.
Bunkers full.
Anchors and cables, complete and stowed.
No provisions or stores on board.
Engineers' stores, half on board.
Hull complete.
The mean deviation in 15 feet for a shift of 25 tons through 36 feet is
lOfk inches = 10*312 inches.
... GM = 2 5 x 3 6 x 15 X 12 =
10312 x 5372
The ship being in an incomplete condition at the time of
the inclining experiment, it was necessary to take an accurate
account of all weights that had to go on board to complete,
with their positions in the ship, together with an account of
all weights that had to be removed, with their positions. The
total weights were then obtained, together with the position of
their final centre of gravity, both in a longitudinal and vertical
direction. For the ship of which the inclining experiment is
given above, it was found that to fully complete her a total
weight of 595 tons had to be placed on board, having its
centre of gravity u feet before the midship ordinate, and 3*05
feet below the designed L.W.L. Also 63 tons of yard plant,
men, etc., had to be removed, with centre of gravity 14 feet
abaft the midship ordinate, and 15 feet above the designed
L.W.L. The centre of buoyancy of the ship at the experi
mental waterline was 10*8 feet abaft the midship ordinate,
and the transverse metacentre at this line was calculated at
314 feet above the designed L.W.L.
We may now calculate the final position of the centre of
gravity of the completed ship as follows, remembering that
in the experimental condition the centre of gravity must be
in the same vertical line as the centre of buoyancy. The
vertical position of G in the experimental condition is found
by subtracting the experimental GM, viz. 2*92 feet, from the
height of the metacentre above the L.W L. as given above,
viz. 3' 1 4 feet.
Conditions of Equilibrium, Transverse Metacentt <?, etc. 125
H
Above
L.W.L.
Below
L.W.L.
Abaft
amidships.
Bel
amids
C
j
ore
hips.
5
Moment.
3^5
Moment.
i
%
Moment.
Moment.
Weight of ship at time"!
of experiment ...J
Weight to go on board)
to complete ... .../
Weight to be takenj
from ship /
5372
595
0'22
Il82
1813
108
58,017
II
6545
5967
63
15
Il82
945
1813
140
58,017
882
6545
5904 237 1813 57,135 6545
237 6,545
1576
50,590
The final position of the centre of gravity of the ship is
therefore
= 0266 feet below the L.W.L.
= 8 '57 feet abaft amidships
the final displacement being 5904 tons.
The mean draught corresponding to the displacement can
be found by the methods we have already dealt with, and corre
sponding to this draught, we can find on the metacentric
diagram the position of the transverse metacentre. In this case
the metacentre was 273 feet above the L.W.L., and conse
quently the value of GM for the completed condition was
273 + 0^266 = 2^996 feet
or say, for all practical purposes, that the transverse metacentric
height in the completed condition was 3 feet.
It is also possible to ascertain what the draughts forward
and aft will be in the completed condition, as we shall see in
the next chapter.
Values of GM, the "Metacentric Height." We
have discussed in this chapter the methods adopted to find
for a given ship the value of the transverse metacentric height
GM. This distance depends upon two things : the position of
G, the centre of gravity of the ship and the position of M, the
126 Theoretical Naval Architecture.
transverse melacentre. The first is dependent on the vertical
distribution of the weights forming the structure and lading 01
the ship, and its position in the ship must vary with differences
in the disposition of the cargo carried. The transverse meta
centre depends solely upon the form of the ship, and its
position can be completely determined for any given draught
of water when we have the sheer drawing of the vessel. There
are two steps to be taken in finding its position for any given
ship floating at a certain waterline.
1. We must find the vertical position of the centre of
buoyancy, the methods adopted being explained in Chapter II.
2. We then find the distance separating the centre of
buoyancy and the transverse metacentre, or BM, as explained
in the present chapter.
By this means we determine the position of M in the ship.
The methods of estimating the position of G, the centre
of gravity for a new ship, will be dealt with separately in
Chapter VI. ; but we have already seen how the position of G
can be determined for a given ship by means of the inclining
experiment. Having thus obtained the position of M and G in
the ship, we get the distance GM, or the metacentric height.
The following table gives the values of the metacentric height
in certain classes of ships. For fuller information reference
must be made to the works quoted at the end of the book.
Type of ship.
Values of GM.
Harbour vessels, as tugs, etc
Modern protected cruisers...
Modern British battleships
Older central citadel armourclads ...
Shallowdraught gunboats for river service
Merchant steamers (varying according to 1
the nature and distribution of the cargo) /
15 to 18 inches
2 to 2j feet
4 to 5 feet
4 to 8 feet
12 feet
I to 3 feet
1 to 3i feet
The amount of metacentric height given to a vessel is based
largely upon experience with successful ships. In order that
a vessel may be " stiff" that is, difficult to incline by external
forces as, for example, by the pressure of the wind on the
Conditions of Equilibrium, Transverse Metacentre, etc. 1 27
sails the metacentric height must be large. This is seen by
reference to the expression for the moment of statical stability
at small angles of inclination from the upright, viz. ^i#''
W X GM sin (see p. 98)
W being the weight of the ship in tons ; being the angle of
inclination, supposed small. This, being the moment tending
to right the ship, is directly dependent on GM. A " crank "
ship is a ship very easily inclined, and in such a ship the
metacentric height is small. For steadiness in a seaway the
metacentric height must be small.
There are thus two opposing conditions to fulfil
j. The metacentric height GM must be enough to enable
the ship to resist inclination by external forces. This is espe
cially the case in sailingships, in order that they may be able
to stand up under canvas without heeling too much. In the
case of the older battleships with short armour belts and
unprotected ends, sufficient metacentric height had to be pro
vided to allow of the ends being riddled, and the consequent
reduction of the moment of inertia of the waterplane.
2. The metacentric height must be moderate enough (if
this can be done consistently with other conditions being
satisfied) to make the vessel steady in a seaway. A ship which
has a very large GM comes back to the upright very suddenly
after being inclined, and consequently a vessel with small
GM is much more comfortable at sea, and, in the case of a
manofwar, affords a much steadier gun platform.
In the case of sailingships, a metacentric height of from
3 to 3^ feet is provided under ordinary conditions of service,
in order to allow the vessel to stand up under her canvas. It is,
however, quite possible that, when loaded with homogeneous
cargoes, as wool, etc., this amount cannot be obtained, on
account of the centre of gravity of the cargo being higl) up in
the ship. In this case, it would be advisable to take in water
or other ballast in order to lower the centre of gravity, and
thus increase the metacentric height
In merchant steamers the conditions continually vary on
account of the varying nature and distribution of the cargo
128 Theoretical Naval Architecture.
carried, and it is probable that a GM of i foot should be the
minimum provided when carrying a homogeneous cargo (con
sistently with satisfactory stability being obtained at large
inclinations). 1 There are, however, cases on record of vessels
going long voyages with a metacentric height of less than i
foot, and being reported as comfortable and seaworthy. Mr.
Denny (Transactions of the Institution of Naval Architects ',
1896) mentioned a case of a merchant steamer, 320 feet long
^'carrying a homogeneous cargo), which sailed habitually with
a metacentric height of 0*6 of a foot, the captain reporting her
behaviour as admirable in a seaway, and in every way com
fortable and safe.
It is the practice of one large steamship company to lay
down that the metacentric height in the loaded condition is no
greater than is required to secure that the metacentric height
in the light condition is not negative.
Effect on Initial Stability due to the Presence of
Free Water in a Ship. On reference to p. 123, where the
inclining experiment for obtaining the vertical position of the
centre of gravity of a ship is explained, it will be noticed that
special attention is drawn to the necessity for ascertaining
that no free water is allowed to remain in the ship while the
experiment is being performed. By free water is meant water
having a free surface. In the case of the boilers, for instance,
they should either be emptied or run up quite full. We now
proceed to ascertain the necessity for taking this precaution.
If a compartment, such as a ballast tank in the double bottom,
or a boiler, is run up quite full, it is evident that the water will
have precisely the same effect on the ship as if it were a solid
body having the same weight and position of its centre of
gravity as the water, and this can be allowed for with very
little difficulty. Suppose, however, that we have on board in
a compartment, such as a ballast tank in the double bottom, a
quantity of water, and the water does not completely fill the
1 Mr. Pescod, before the NorthEast Coast Institution of Engineers and
Shipbuilders, 1903, dealt with the minimum GM for small vessels. He
there states that it is generally recognized that the GM of cargo vessels
should not be less than O'8 foot provided that a righting arm of like amount
is obtained at 30 to 40 degrees.
Conditions of Equilibrium, Transverse Metacentre^ etc. 129
tank, but has a free surface, as wl^ Fig. 60.! If the ship is
heeled over to a small angle 0, the water in the tank must
adjust itself so that its surface w'l' is parallel to the level water
line W'L'. Let the volume of either of the small wedges wsw',
1st be z> > and g> g 1 the positions of the ircentres of gravity, b, b 1
being the centres of gravity of the whole volume of water in
the upright and inclined positions respectively. Then, if V
be the total volume of water in the tank, we have
V X bV = v Q X gg'
and bb' = =^ X XS*
and bit is parallel to gg.
found the moment of
transference of the
wedges WSW, LSL',
in Fig. 45, we can find
the moment of trans
ference of the small
wedges wsw', /j/,viz.
^o X gg' = i X B
where i is the moment
of inertia of the free
surface of the water in
the tank about a fore
andaft axis through s ;
and 9 is the circular
measure of the angle
of inclination.
Now, in precisely the same way as we
FIG. 60.
Substituting this value for v x gg', we have
i X
Draw the new vertical through ', meeting the middle line in
m\ then
bb' = bmxQ
1 Fig. 60 is drawn out of proportion for the sake of clearness.
J 3 Theoretical Naval Architecture.
and consequently
bm X =
and bm = ^r
Now, if the water were solid its centre of gravity would be
at b both in the upright and inclined conditions, but the weight
of the water now acts through the point b' in the line b'm, and
its effect on the ship is just the same as if it were a solid
weight concentrated at the point m. So that, although b is
the actual centre of gravity of the water, its effect on the ship,
when inclined through ever so small an angle, is the same as
though it were at the point m t and in consequence of this the
point m is termed the virtual centre of gravity of the water. 1
This may be made clearer by the following illustrations :
1. Suppose that one instant the water is solid, with its
centre of gravity at b, and the following instant it became liquid.
Then, for small angles of inclination, its effect on the ship would
be the same as if we had raised its weight through a vertical
distance bm from its actual to its virtual centre of gravity.
2. Imagine a pendulum suspended at m t with its bob at b.
On the ship being inclined to the small angle 0,the pendulum
will take up the position mb\ and this corresponds exactly to
the action of the water.
We thus see that the centre of gravity of the ship cannot be
regarded as being at G, but as having risen to GO, and if W be
the weight of water in tons = f (the water being supposed
salt), we have
W X GG = W X bm
and therefore
GG = V Xm = x bm ( v = volume of displacement)
1 See a paper by Mr. W. Hok, at the Institution of Naval Architects,
1895, on " The Transverse Stability of Floating Vessels containing Liquids,
with Special Reference to Ships carrying Oil in Bulk." See also a paper
in the "Transactions of the Institution of Engineers and Shipbuilders in
Scotland for 1889," by the late Professor Jenkins, on the stability of vessels
carrying oil in bulk.
Conditions of Equilibrium, Transverse Metacentre, etc. 131
But we have seen that
bm = ==
and therefore
GG, = Y X ^ = 1
The new moment of stability at the angle is
W X G M x sin B = W x (GM  GG ) sin 6
= WX
the metacentric height being reduced by the simple expres
sion ==. We notice here that the amount of water does not
affect the result, but only the moment of inertia of the free
surface. The necessity for the precaution of clearing all free
water out of a ship on inclining is now apparent. A small
quantity of water will have as much effect on the position of
the centre of gravity, and therefore on the trustworthiness of
the result obtained, as a large quantity of water, provided it
has the same form of free surface. If a small quantity of
water has a large free surface, it will have more effect than
a very large quantity of water having a smaller free surface.
If the liquid contained is other than the water the vessel is
P* i
floating in, the loss of metacentric height is ^, where p is the
specific gravity of liquid compared with outside water, and V
the total volume of displacement.
Example. A vessel has a compartment of the double bottom at the
middle line, 60 feet long and 30 feet broad, partially filled with salt water.
The total displacement is 9100 tons, and centre of gravity of the ship and
water is o  26 feet below the waterline. Find the loss of metacentric
height due to the water having a free surface.
We have here given the position of the centre of gravity of the ship and
the water. The rise of this centre of gravity due to the mobility of the
water is, using the above notation
i
V
and / = ^(60 X 30) X (30)*
= 5 X (30)'
Since the free surface is a rectangle 60 feet long and 30 broad
and V = 9100 x 35 cubic feet
therefore the loss in metacentric height = = = 0424 feet
9100 x 35
132 Theoretical Naval Architecture.
Met acentric Diagrams for Simple Figures. i. A
rectangular box. This is dealt with on p. 116.
B 2
BM = Ta "n B = breadth D = draught,
D B a
and M from base is  h ^ jj
By the methods of the calculus this is found to be a minimum
when D 2 = \ . B 2 , i.e. when M is in the W.L. or where it crosses
the 45 line.
The M curve is a hyperbola referred to the vertical at zero
draught, and the C.B. line as axes, having the equation
the axes being asymptotes.
2. A vessel with a triangular section, vertex down. This is
dealt with on p. 117, where it is seen that the M curve is
straight
BM = J.^. M from base = f .D + J.jj
M from base B 2
/.  g  f + f jJJ I + I tan a = constant
Tt
Pj = tan a, where a is the semivertical angle
i.e. M curve is a straight line, making an angle 0, with the
base line such that tan = (i + tan 2 a).
3. Vessel with parabolic section. A parabola has the equa
tion referred to axes at the vertex, y 2 = 4ax, i.e. for x draught
breadth at waterline is 2y = 4* ax (Fig. 6oA).
The C.G. of a parabola is f the depth, so that the C.B.
locus is a straight line making an angle with the base of
tan^d). Area of parabola = f . circumscribing rectangle.
BM = f.2.lf = * ' D = = C nStant
i.e. the locus of metacentres in metacentric diagram is straight
and parallel to the C.B. locus.
Conditions of Equilibrium, Transverse Metacentre, etc. 133
4. Vessel with circular section (Fig. 6oB). In this case the
metacentre is always at the centre, so that the M curve is a
straight line at mid depth.
FIG. 6oA.
FIG. 6oB.
The B curve is a flat curve starting at an angle with the
base, such that = tan" 1 ^), since there the circle may be
regarded as a parabola.
For mid depth B below W.L. is  (a being radius), and
3^
the inclination of tangent is an angle a such that tan a = '54
by the formula given on p. 118.
When completely immersed the curve finishes as a tangent
to the M curve.
Curves of Buoyancy, etc. The surface of biioyancy for a
given displacement is the surface traced out by the centre of
buoyancy as the vessel takes up all possible positions while
maintaining that displacement.
The surface of flotation is the surface traced out by the centre
of flotation under the same conditions.
The curve of buoyancy is the curve traced out on the transverse
vertical plane by the projection of the centre of buoyancy as the
ship is continually revolved about a longitudinal axis fixed in
direction while maintaining the same displacement. This curve is
also termed an isovol.
The curve of flotation is the curve traced out by the projection
of the centre of flotation under the same conditions.
134 Theoretical Naval Architecture.
The prometacentre is the intersection of any two consecutive
lines of action of buoyancy, as M' in Fig. 6oE. When consecutive
lines do not intersect the prometacentre is the intersection of one
of them with the common perpendicular. For a condition of
equilibrium this intersection of consecutive lines of buoyancy is
the metacentre.
The metacentric is the locus of prometacentres.
The following are definitions of various sorts of equilibrium:
(1) Rotation in a given direction only.
(a) Stable equilibrium for a given direction of inclination
when, on being slightly displaced in that direction
from its position of rest, the vessel tends, on being
released, to go back to that position.
() Unstable equilibrium is as (a), only that the vessel moves
further from the position of rest.
(c) Indifferent or unstable equilibrium the vessel neither
tends to return to or to go further from the position of
rest.
(rf) Mixed Equilibrium if stable for one direction of in
clination and unstable for the opposite direction.
(2) Rotation in all directions.
(a) Absolute equilibrium when only stable or unstable for
any direction of inclination.
(b) Relative stability when stable in some directions and
unstable in others.
Thus, in a ship
(i) If the C.G. is below the transverse metacentre M T , she is
absolutely stable,
(ii) If the C.G. is above the longitudinal metacentre M L ,
she is absolutely unstable.
(iii) If the C.G. is between M T and M L , she has relative
stability, being stable for longitudinal inclinations and
unstable for transverse inclinations.
The above definitions are well illustrated by a floating cube of
s.g. i.
(a) When floating with a face horizontal, the cube is
absolutely unstable.
(b] With one corner downwards, the cube has absolute
stability.
In going from one point on the surface of buoyancy to the con
secutive point, B to B', BB' = ^ y^% and BB' is parallel to g^.
Hence, in the limit BB' is parallel to the waterplane, so that the
Conditions of Equilibrium, Transverse Metacentre, etc. 135
tangent plane at any point to the surface of buoyancy is parallel to
the corresponding waterplane, and the normal to it through the
point of contact gives the line of action of buoyancy. The surface
must be wholly convex to the tangent plane or wholly concave to
some interior point. Similar reasoning will also apply to the curve
of buoyancy.
For a position of equilibrium, the line of action of the buoyancy
must pass through the C.G. ; therefore, the number of positions of
equilibrium that a body can take up is equal to the number of
normals that can be drawn from the C.G. to the surface of buoyancy.
For a given direction of inclination the number of positions of equi
librium equals the number of normals that can be drawn from the
C. G. to the curve of buoyancy.
When BG thus drawn is a minimum, the equilibrium is stable.
When BG thus drawn is a maximum, the equilibrium is
unstable ;
for the stability is the same as that of the curve of buoyancy
rolled along a smooth horizontal plane, the weight being concen
trated at the C.G. In moving from one position of equilibrium to
another, if the C.G. has to be raised we have stable equilibrium,
i.e. B'G > BG. If unstable, similarly B'G < BG.
The centre of curvature of the surface of buoyancy is what we
have termed the prometacentre, and the radius of curvature is
given by R =^ where I is the moment of inertia of the water
plane about an axis through its C.G. perpendicular to the plane of
rotation, and V is the volume of
displacement. This is proved
exactly as in Chap. III. for the
upright BM.
Lederfs theorem for the radius
of curvature of the curve of
flotation.
In Fig. 6oc WL and W'L' are
consecutive waterlines for the
upright ivl, iv'l', when inclined to
a small angle, the increment of
displacement being AV. Then
when inclined the buoyancy V
acts through M, and that of AV through O, the centre of curva
ture of the curve of flotation. B and B' are the upright C.B.'s
and M' the metacentre for the waterline W'L', and V + AV acts
through M' for a small inclination.
i
FIG. 6oc.
136 Theoretical Naval Architecture.
Taking moments about C, the C.G. of the layer AV, we
have
(V x CM) + (AV x CO) = (V + AV)CM'
or V(BM  BC) + (AV x CO) = (V + AV)(B'M'  B'C)
now V x BC = (V + AV)B'C
so that (V x BM) + (AV x CO) = (V + AV)B'M'
or I + (AV x CO) = I + Al
d\
i.e. OC = jr.. in the Limit, which is
the expression for the radius of curvature of the curve of flotation
usually called r.
It can be readily shown that if a weight be added at the point
O the moment of initial stability is not changed. For ordinary
ships parallelsided at the waterline d\ is zero or practically so, so
that O is in the waterline. We may therefore say that, generally
speaking, if a weight is added above the waterline it will diminish
the stability ; if added at the waterline there is no change in
the stability ; if added below the waterline the stability is
increased.
Examples. (i.) r for a body of rectangular section, r o.
(ii.) r for a body of triangular equilateral section, angle 20.
(a) corner downwards, r d tan 2 (d = draught).
(b) corner upwards, r = c tan 2 (c being dis
tance of waterline from vertex).
(iii.) r for a circular section, radius a, r = a cos 6.
(26 being angle subtended at the centre by the waterline.)
(iv.) Show that an added weight to keep the metacentric
height constant should be placed the same distance
from G as O is distant from M.
(v.) In Example (ii.) () above, if tan 6 = f and depth is 40'
then if the draught is less than 14*4 ft. a small addition
of ballast to the base of the triangle will make the
body less stable, but at greater draughts the stability
increases with the addition of ballast.
GEOMETRY OF THE METACENTRIC DIAGRAM.
r. Tangent to the curve of C.B. In Fig. 6oD, let be the
inclination of this locus to the horizontal at waterline WL.
Then for an increment of displacement AV and of draught by
the C.B. will rise an amount ^ . AV = ' . Ay.
Conditions of Equilibrium, Transverse Metacentre, etc. 1 37
tan e = IL s  e 2O? = A h W here A = area of waterplane,
h = C.B. below waterline,
V = volume displacement.
w
w
V+AV
Mr
FIG. 600.
Examples. (i.) for a boxshaped vessel, tan 6 = O'5
(ii.) for a triangular section, tan = O'66.
(iii.) for a circular section at) , Q
half depth } tan = o 54.
For ordinary ships it is found that tan Q = 0*55 about.
2. Tangent to the curve of metacentres. The increment of
volume AV, for a small inclination, has its line of action through O,
the centre of curvature of the curve of flotation. Let OM = k
then (V + AV)MM' = AV x OM . = AV x k (Fig. 6oc)
MM' =
V
Ay.
If $ be the angle the tangent to the M curve makes with the
horizontal, then
rise or fall of M A . k .
tan <}> =   = 
in the limit
If k = o, then tan </> = o and the M curve is horizontal, i.e. when
the M curve is horizontal, the points M and O coincide. This is
otherwise obvious, as the added buoyancy will act through M, which
is therefore fixed in height for a small increment of draught. In a
boxshaped ship the M curve is horizontal when D 2 = \ B 2 .
Coordinates of the Centre of Buoyancy referred to axes through
the itpright C.B.
In Fig. 6oE, x and y are the coordinates of B, the C.B. at
angle 0. For an increment of angle dQ, B' is the new C.B. and
133
Theoretical Naval Architecture.
x + dx, y 4 tf^the coordinates, BM', B'M' the normals at B and B'
intersect in M' the prometacentre, and BM' = B'M' = R, and
R = ^ . BB' = R . dQ, and dx  BB' . cos 6, dy = BB' . sin 0,
.*. dx = R . cos e . d9 dy = R . sin . dQ
and x = JR . cos 6 . d9 y  /R . sin e . d9.
Curves of R cos 6, R . sin can be drawn on a 6 base and integrated
up to the various angles. Thus, x and_x can be obtained and so
the curve of buoyancy drawn in. The righting arm at angle is
given by GZ = .r.cos 6 + y sin 9 B G.sin e.
FIG. 6oE.
FIG. 6oF.
This is the French method of calculating stability due to
M. Reech
For a box so long as wall sided R = B M . sec 2 6
so that x JB M . sec 3 & cos 6 dQ B M . tan 9
y = JB M sec 3 B . sin 6 dQ
= j*B M . sec 2 6 tan d6  B M . tan 2
This is the solution of question 35 in the Appendix.
Question 36 is solved as follows :
G' the new C.G. will lie on BM' and GG' = ^^
GG' cos 6 = x cos 6 + y sin B B G sin 6
GG' x +y . tan B G . tan 6
= B M . tan + JB M . tan 3 e  B G . tan 6
= GM . tan e + *B M . tan 3 6
Also
GG' =
w X d
W
f
Conditions of Equilibrium, Transverse Metacentre, etc. 1 39
STABILITY OF A WALLSIDED VESSEL,
, v x hh 1
In Fig. 6oF, BR = =
v ~ \ y 2  tan e
hh' f the projection of aa on to W'L'
= ^ the projection of SL 4 La on to W'L'
= ^ (_y . cos + ^ .y . tan . sin 0)
:= = BM . sin 0(i + tan 2 0) taking a unit length
= BR also.
GZ = BR  EG. sin
= BM sin + BM . tan 2 sin  BG sin
= sin 0(GM + BM . tan 2 0).
This can be used to construct the curve of stability (see Chap.
V.) so far as the ship is wallsided above and below water, and
can be used to check the cross curve at 15 say obtained by the
Integrator.
This formula may be used to determine the angle to which a
ship with negative metacentric height will loll over, for GZ will
then be zero, and we have
and the metacentric height when at the angle is 2 . (See
example 37 in Appendix.)
EXAMPLES TO CHAPTER III.
1. Find the circular measure of 5 , ioj, 15!.
Ans. 009599; 017889; 027489.
2. Show that sin 10 is onehalf per cent, less in value than the circular
measure of 10, and that tan 10 is one per cent, greater in value than the
circular measure of 10.
3. A cylinder weighing 500 Ibs., whose centre of gravity is 2 feet from
the axis, is placed on a smooth table and takes up a position of stable
equilibrium. It is rolled along parallel to itself through an angle of 60.
What will be the tendency then to return to the original position ?
Ans. 866 footlbs.
4. Find the moment of inertia about the longest axis through the centre
of gravity of a figure formed of a square of side 20, having a semicircle at
each end.
140 Theoretical Naval Architecture.
5. Find the moment of inertia of a square of side 2a about a diagonal.
Am. \ a\
6. A square has a similar square cut out of its centre such that' the
moment of inertia (about a line through the centre parallel to one side) of
the small square and of the portion remaining is the same. What pro
portion of the area of the original square is cut out ?
Ans. 071 nearly.
7. A vessel of rectangular crosssection throughout floats at a constant
draught of 10 feet, and has its centre of gravity in the load water plane.
The successive halfordinates of the load waterplane in feet are o'5, 6, 12,
16, 15, 9, o; and the common interval 20 feet. Find the transverse
metacentric height.
Ans. 8 inches.
8. A log of fir, specific gravity 0*5, is 12 feet long, and the section is
2 feet square. What is its transverse metacentric height when floating in
stable equilibrium in fresh water ?
Ans. o  47 foot.
9. The semiordinates of a waterplane 34 feet apart are 0*4, 13*7,
25*4, 321, 346, 350, 349, 342, 321, 239, 69 feet respectively. Find
its moment of inertia about the centre line.
Ans. 6,012,862.
10. The semiordinates of the load waterplane of a vessel are o, 3*35,
641, 863, 993, 1044, I0'37, 9'94 896, 716, and 25 feet respectively.
These ordinates being 21 feet apart, find
(1) The tons per inch immersion.
(2) The distance between the centre of buoyancy and the transverse
metacentre, the load displacement being 484 tons.
Ans. (i) 773 tons; (2) 5*2 feet nearly.
11. The semiordinates, l6'6 feet apart, of a vessel's waterplane are
02, 23, 64, 99, 123, 135, 138, 137, 128, 106, 64, i'9, 02 feet
respectively, and the displacement up to this waterplane is 220 tons. Find
the length of the transverse BM.
Ans. 2O'6 feet.
12. A vessel of 613 tons displacement was inclined by moving 30 cwt.
of rivets across the deck through a distance of 22' 6". The end of a plumb
line 10 feet long moved through 2\ inches. What was the metacentric
height at the time of the experiment ?
Ans. 2 93 feet.
13. The semiordinates of a ship's waterplane 35 feet apart are, com
mencing from forward, 0*4, 7*12, 15*28, 2i'8, 2562, 269, 2632, 2442,
20 '8, I5'I5? 6 39 feet respectively. There is an after appendage of 116
square feet, with its centre of gravity 180 feet abaft the midship ordinate.
Find
(1) The area of the waterplane.
(2) The tons per inch immersion.
(3) The distance of the centre of flotation abaft amidships.
(4) The position of the transverse metacentre above the L.W.L., taking
the displacement up to the above line as 5372 tons, and the
centre of buoyancy of this displacement 8'6i feet below the
L.W.L.
Ans. (I) 1 3, 292 square feet; (2) 316 tons; (3) 14*65 feet ; (4) 334
feet.
14. A ship displacing 9972 tons is inclined by moving 40 tons 54 feet
Conditions of Equilibrium, Transverse Metacentre, etc. 1 4 1
across the deck, and a mean deviation of 9^ inches is obtained by pendulums
15 feet long. Find the metacentric height at the time of the operation.
Ans. 4* 1 8 feet.
15. A ship weighing 10,333 tons was inclined by shifting 40 tons 52
feet across the deck. The tangent of the angle of inclination caused was
found to be 0*05. If the transverse metacentre was 475 feet above the
designed L. W.L., what was the position of the centre of gravity of the ship
at the time of the experiment ?
Ans. 073 foot above the L.W.L.
1 6. A vessel of 26 feet draught has the moment of inertia of the L. W. P.
about a longitudinal axis through its centre of gravity 6, 500,000 in foot
units. The area of the L.W.P. is 20,000 square feet, the volume of dis
placement 400,000 cubic feet, and the centre of gravity of the ship may be
taken in the L.W.P. Approximate to the metacentric height.
Ans. 5^ feet.
17. Prove the rule given on p. 62 for the distance of the centre of
gravity of a semicircle of radius a from the diameter, viz. ^a, by finding
the transverse BM of a pontoon of circular section floating with its axis in
the surface of the water.
(M in this case is in the centre of section.)
1 8. Take a body shaped as in Kirk's analysis, p. 84, of length 140
feet ; length of parallel middle body, loo feet'; extreme breadth, 30 feet ;
draught, 12 feet. Find the transverse BM.
Ans. 57 feet.
19. A vessel of 1792 tons displacement is inclined by shifting 5 tons
already on board transversely across the deck through 20 feet. The end
of a plumbline 15 feet long moves through 5J inches. Determine the
metacentric height at the time of the experiment.
Ans. 191 feet.
20. A vessel of displacement 1722 tons is inclined by shifting 6 tons of
ballast across the deck through 22 \ feet. A mean deviation of loj inches
is obtained with pendulums 15 feet long. The transverse metacentre is
15*28 feet above the keel. Find the position of the centre of gravity of the
ship with reference to the keel.
Ans. 13*95 f et
21. The ship in the previous question has 169 tons to go on board at
lo feet above keel, and 32 tons to come out at 20 feet above keel. Find
the metacentric height when completed, the transverse metacentre at the
displacement of 1859 tons being 153 feet above keel.
Ans. i 8 feet.
22. A vessel of 7000 tons displacement has a weight of 30 tons moved
transversely across the deck through a distance of 50 feet, and a plumbbob
hung down a hatchway shows a deviation of 12 inches in 15 feet. What
was the metacentric height at the time of the operation ?
Ans. 3 '2 1 feet.
23. A box is 200 feet long, 30 feet broad, and weighs 2000 tons. Find
the height of the transverse metacentre above the bottom when the box is
floating in salt water on an even keel. Ans. 12 '26 feet.
24. Show that for a rectangular box floating at a uniform draught of d
feet, the breadth being 12 feet, the distance of the transverse metacentre
above the bottom is given by 24 feet, and thus the transverse meta
centre is in the waterline when the draught is 49 feet.
142 Theoretical Naval Architecture.
25. A floating body has a constant triangular section. If the breadth
at the waterline is A/ 2 times the draught, show that the curve of metacentres
in the metacentric diagram lies along the line drawn from zero draught at
45 to the horizontal, and therefore the metacentre is in the waterline for
all draughts.
26. A floating body has a square section with one side horizontal.
Show that the transverse metacentre lies above the centre of the square
so long as the draught does not much exceed 21 per cent, of the depth of
the square. Also show that as the draught gets beyond 21 per cent, of the
depth, the metacentre falls below the centre and remains below until
the draught reaches 79 per cent, of the depth ; it then rises again above
the centre of the square, and continues to rise as long as any part of the
square is out of the water.
(This may be done by constructing a metacentric diagram, or by using the
methods of algebra, in which case a quadratic equation has to be solved.)
27. Show that a square log of timber of 12 inches side, 10 feet long, and
weighing 320 Ibs., must be loaded so that its centre of gravity is more than
I inch below the centre in order that it may float with a side horizontal
in water of which 35 cubic feet weigh I ton.
28. A prismatic vessel is 70 feet long. The section is formed at the
lower part by an isosceles triangle, vertex downwards, the base being 20
feet, and the height 5 feet ; above this is a rectangle 20 feet wide and 5 feet
high. Construct to scale the metacentric diagram for all drafts.
29. A vessel's load waterplane is 380 feet long, and 75 feet broad, and
its moment of inertia in footunits about the centre line works out to
8,000,000 about. State whether you consider this a reasonable result to
obtain, the waterplane not being very fine.
TJJ
30. Find the value of the coefficient a in the formula BM = a
referred to on p. in, for floating bodies having the following sections
throughout their length :
(a) Rectangular crosssection.
(6) Triangular crosssection, vertex down.
(c) Verticalsided for one half the draught, the lower half of the section
being in the form of a triangle.
Arts, (a) 008; (6) ci6 ; (c) o'li.
For ordinary ships the value of a will lie between the first and last of these.
31. A lighter in the form of a box is 100 feet long, 20 feet broad, and
floats at a constant draught of 4 feet. The metacentric height when empty
is 6 feet. Two bulkheads are built 10 feet from either end. Show that a
small quantity of water introduced into the central compartment will render
the lighter unstable in the upright condition.
32. At one time, in ships which were found to possess insufficient sta
bility, girdling was secured to the ship in the neighbourhood of the water
line. Indicate how far the stability would be influenced by this means.
33. A floating body has a constant triangular section. If the breadth
at the waterline is equal to the draught, show that the locus of metacentres
in the metacentric diagram makes an angle with the horizontal of about 40.
34. A cylinder is placed into water with its axis vertical. Show that if
the centre of gravity is in the waterplane, the cylinder will float upright if
the radius r the draught is greater than */2.
35. In a wholly submerged body show that for stable equilibrium the
centre of gravity must lie below the centre of buoyancy.
36. A floating body has a constant triangular section, vertex down
wards, and has a constant draught of 12 feet, the breadth at the waterline
Conditions of Equilibrium, Transverse Metacentre, etc. 143
being 24 feet. The keel just touches a quantity of mud, specific gravity 2.
The waterlevel now falls 6 feet : find the amount by which the meta
centric height is diminished due to this. 1 Ans. 2f feet about.
37. A floating body of circular section 6 feet in diameter has a meta
centric height of I "27 feet. Show that the centre of buoyancy and centre
of gravity coincide, when the body is floating with the axis in the surface.
38. It is desired to increase the metacentric height of a vessel which is
being taken in hand for a complete overhaul. Discuss the three following
methods of doing this, assuming the ship has a metacentric diagram as in
Fig. 56, the extreme load draught being 15 feet :
(1) Placing ballast in the bottom.
(2) Removing top weight.
(3) Placing a girdling round the ship in the neighbourhood of the
waterline.
39. Show that the angle in Fig. 56 is between 29 and 30 for a
vessel whose coefficient of L.W.P. is 075, and whose block coefficient
of displacement is 055. In any case, if these coefficients are denoted by
n and k respectively, show that tan = \ + ^ approximately (use Mor
rish's formula, p. 65).
40. From the following information construct the metacentric diagram,
using a scale of inch = I foot, and state the metacentric height and
draught in the three conditions given.
Draught.
Displacement
in tons.
Tons per
inch.
C.B. below
i9foot WL.
BM.
2i ; 9 ;:
19' o"
5256
4383
27I
2648
625'
78'
885'
104'
16' 3"
3527
25'37
9'35'
122'
13' 6"
2714
2384
109'
145'
(1) Deep load 1000 tons coal 5030 tons, C.G. o'3 feet below 19' WL.
(2) Normal load 400 4430 035
(3) Light condition 3915 ,, 0*3 above ,,
Ans. (i) 29', 21' of'; (2) 295', 19' if" ; (3) 2' 4 ', 1/6"
41. A weight of 10 tons is shifted 40 ft. transversely across the deck
of a vessel having a compartment partially filled with salt water, the free
surface of this water being 25 ft. long by 50 ft. uniform breadth. Calculate
the heel in degrees, having given displacement of vessel 8000 tons, C.G. of
ship and contained water 15 ft. above keel. Transverse M. i6 ft. above
keel. (Durham B.Sc. 1910).
The actual GM is i feet, but the virtual GM is less than this by the
amount ^r, where * is the moment of inertia of free water surface about
a longitudinal axis through its C.G.
* = & 2 S.So 50* V = 8000 X 35
so that ^ = 093 ft. The virtual GM is therefore 057 ft., 9 being angle
of heel.
W x GM X sin = w X d
or sin =
10 X 40
= 0088
8000 X 057
and = 5 degrees.
1 This example is worked out at the end of the Appendix.
CHAPTER IV.
LONGITUDINAL METACENTRE, LONGITUDINAL BM,
CHANGE OF TRIM.
Longitudinal Metacentre. We now have to deal with
inclinations in a foreandaft or longitudinal direction. We
do not have the same difficulty in fixing on the foreandaft
position of the centre of gravity of a ship as we have in fixing
its vertical position, because we know that if a ship is floating
steadily at a given waterline, the centre of gravity must be in
the same vertical line as the centre of buoyancy, by the con
ditions of equilibrium laid down on p. 93. It is simply a
matter of calculation to find the longitudinal position of the
centre of buoyancy of a ship when floating at a certain water
line, if we have the form of the ship given, and thus the fore
andaft position of the centre of gravity is determined.
We have already dealt with the inclination of a ship in a
transverse direction, caused by shifting weights already on
board across the deck ; and in a precisely similar manner we
can incline a ship in a longitudinal or foreandaft direction by
shifting weights along the deck in the line of the keel. The
trim of a ship is the difference between the draughts of water
forward and aft. Thus a ship designed to float at a draught
forward of 12 feet, and a draft aft of 15 feet, is said to trim 3 feet
by the stern.
We have, on p. 97, considered the definition of the trans
verse metacentre, and the definition of the longitudinal meta
centre is precisely analogous.
For a given waterline WL of a vessel, let B be the centre
of buoyancy (see Fig. 61), and BM the vertical through it.
Longitudinal Metacentre, Longitudinal BM, etc. 145
Suppose the trim of the vessel to change slightly, 1 the vessel
retaining the same volume of displacement, B' being the new
centre of buoyancy, and B'M the vertical through it, meeting
FIG. 61.
BM in M. Then the point M is termed the longitudinal
metacentre.
The distance between G, the centre of gravity of the ship,
and M, the longitudinal metacentre, is termed the longitudinal
metacentric height.
Formula for finding the Distance of the Longi
tudinal Metacentre above the Centre of Buoyancy.
Let Fig. 62 represent the profile of a ship floating at the water
line WL', the original waterline being WL. The original
trim was AW  BL ; the new trim is AW  BL'. The change
of trim is
(AW  BL)  (AW  BL') = WW + LL'
i.e. the change of trim is the sum of the changes of draughts
forward and aft. This change, we may suppose, has been
caused by the shifting of weights from aft to forward. The
inclination being regarded as small, and the displacement
remaining constant, the line of intersection of the waterplanes
WL, W'L' must pass through the centre of gravity of the water
plane WL, or, as we have termed it, the centre of flotation,
in accordance with the principle laid down on p. 98. This
centre of flotation will usually be abaft the middle of length,
and this introduces a complication which makes the calculation
for the longitudinal metacentre more difficult than the corre
1 Much exaggerated in the figure.
146
Theoretical Naval Architecture.
spending calculation for the transverse metacentre. In this
latter case, it will be remembered that the centre of flotation is
in the middle line of the waterplane.
FIG. '62.
In Fig. 62
Let B be the centre of buoyancy when floating at the
waterline WL ;
B', the centre of buoyancy when floating at the water
line WL' ;
FF, the intersection of the waterplanes WL, WL' ;
v, the volume of either the immersed wedge FLL' or
the emerged wedge FWW ;
, ^, the centres of gravity of the wedges WFW, LFL
respectively ;
V, the volume of displacement in cubic feet ;
0, the angle between the waterlines WL, WL', which
is the same as the angle between BM and B'M
(this angle is supposed very small).
We have, using the principle laid down on p. 100
v X = V v BB'
Longitudinal Metacentre, Longitudinal BM, etc. 147
, r PR'
or BB =
But BB' = BM x 6 (0 is in circular measure)
The part of this expression that we do not know is v X gg\
or the moment of transference of the wedges. At P take a
small transverse slice of the wedge FLL', of breadth in a fore
andaft direction, dx\ length across, 2y; and distance from
F, x. Then the depth of the slice is
x x
and the volume is 2y x xO X dx
This is an elementary volume, analogous to the elementary
area y . dx used in finding a large area. The moment of this
elementary volume about the transverse line FF is
2yx . . dx X x
or 2yx* .6 .dx
If we summed all such moments as this for the length FL,
we should get the moment v X F^', and for the length FW,
v X F, or for the whole length, v X gg' j therefore, using our
ordinary notation
X gg = f2yx* .O.dx
= 26jyx*.dx (6 being constant)
We therefore have
or
Referring to p. 103, it will be seen that we denned the
moment of inertia of an area about a given axis as
JWA X/
where dA. is a small elementary area ;
y its distance from the given axis.
Consider, now, the expression obtained, 2fyx? . dx. The
elementary area is 2y . dx, and x is its distance from a
148
Theoretical Naval Architecture.
transverse axis passing through the centre of flotation. We
may therefore say
where I is the moment of inertia of the waterplane about a
transverse axis passing through the centre of flotation. It will
be seen at once that this is the same form of expression as for
the transverse BM.
The method usually adopted for finding the moment of
inertia of a waterplane about a transverse axis through the
centre of flotation is as follows * :
We first find the moment of inertia about the ordinary
midship ordinate. If we call this I, and y the distance of the
centre of flotation from the midship ordinate, we have, using
the principle given on p. 104
I = I + A/
or I = I  A/
The method actually adopted in practice will be best under
stood by working the following example.
Numbers
of
ordinates.
Semi
ordinates
ofL.W.P.
Simpson's
multi
pliers.
Products
for area.
Multi
pliers for
moment.
Products
for
moment.^
Multi
pliers for
moment
of inertia.
Products
for
moment
of inertia.
I
O'O
*
O'O
5
oo
5
O'O
I*
i'37
2
274
4i
1233
4i
5 5 '49
2
267
ii
401
4
1604
4
6416
3
487
4
1948
3
5844
3
i75'32
4
631
2
1262
2
2524
2
5048
5
685
4
2740
I
2740
I
2740
6
721
2
1442
o
I3945
O
7
7*15
4
2860
I
2860
I
2860
g
687
2
1374
2
2748
2
5496
9
6'33
4
2532
3
75^6
3
22788
10
508
I*
762
4
3048
4
12192
10*
356
2
712
4*
3204
4
14418
ii
071
i
o'35
5
175
5
875
16342 19631 9S9'i4
= Si I 39'45 = S,
5686 =S 2
1 This calculation for the L.W.P. is usually performed on the displace
ment sheet.
Longitudinal Metacentre, Longitudinal BM y etc. 149
In column 2 of the table are given the lengths of semi
ordinates of a load waterplane corresponding to the numbers
of the ordinates in column i. The ordinates are 7*1 feet
apart. It is required to find the longitudinal BM, the dis
placement being 916 tons in salt water.
The distance apart of the ordinates being 7*1 feet, we have
Area = 163*42 X (J X 7'i) X 2
= 7 7 3' 5 square feet
Distance of centre of gravity of 1 __ 56*86 X 7'i
waterplane abaft No. 6 ordinate J " 163*42 2 4 ^
(the stations are numbered from forward).
The calculation up to now has been the ordinary one
for finding the area and position of the centre of gravity.
Column 4 is the calculation indicated by the formula
Area = 2Jy . dx
Column 6 is the calculation indicated by the formula
Moment = 2Jyx . dx
It will be remembered that in column 5 we do not put
down the actual distances of the ordinates from No. 6 ordinate,
but the number of intervals away; the distance apart of the
ordinates being introduced at the end. By this means the
result is obtained with much less labour than if column 5
contained the actual distances. The formula we have for the
moment of inertia is 2Jy . x* . dx. We follow a similar process
to that indicated above ; we do not multiply the ordinates by
the square of the actual distances, but by the square of the
number of intervals away, leaving to the end the multiplication
by the square of the interval. Thus for ordinate No. 2 the
actual distance from No. 6 is 4X7*1 = 28 "4 feet. The
square of this is (4) 2 X (7*i) 2 . For ordinate No. 4 the square of
the distance is (2) 2 X (7*i) 2 . The multiplication by (7*i) 2 can
be done at the end. In column 7 is placed the number of
intervals from No. 6, as in column 5 ; and if the products in
column 6 are multiplied successively by the numbers in
column 7, we shall obtain in column 8 the ordinates put
15 Ttteoretical Naval Architecture.
through Simpson's rule, and also multiplied by the square of
the number of intervals from No. 6 ordinate. The whole of
column 8 is added up, giving a result 959*14. To obtain the
moment of inertia about No. 6 .ordinate, this has to be multi
plied as follows :
(a) By onethird the common interval to complete Simp
son's rule, or \ X 7*1.
(b) By the square of the common interval, for the reasons
fully explained above.
(c) By two for both sides.
We therefore have the moment of inertia of the waterplane
about No. 6 ordinate
95914 x ( X 71) X (7'i) 2 X 2 = 228,858
The moment of inertia about a transverse axis through the
centre of flotation will be less than this by considering the
formula I = I + Ay, where I is the value found above about
No. 6 ordinate, and I is the moment of inertia we want. We
found above that the area A = 773*5 square feet, and y = 2*47
feet;
.Mo =228,858 (773*5 X 2 47 2 )
= 224,1391
The displacement up to this waterplane is 91*6 tons, and
the volume of displacement is
91*6 x 35 = 3206 cubic feet
The longitudinal BM =
3206
Approximate Formula for the Height of the Longi
tudinal Metacentre above the Centre of Buoyancy.
The following formula is due to M. J. A. Normand, M.I.N.A., 2
and is found to give exceedingly good results in practice :
Let L be the length on the load waterline in feet ;
B, the breadth amidships in feet ;
1 See note at end of chapter, p. 167.
8 See "Transactions of the Institution of Naval Architects, ' 1882.
Longitudinal Metacentre, Longitudinal BM, etc. 151
V, the volume of displacement in cubic feet ;
A, the area of the load waterplane in square feet.
Then the height of the longitudinal metacentre above the centre
of buoyancy
A 2 XL
H = 00735B1TV
In the example worked above, the breadth amidships was
14*42 feet; and using the formula, we find
H = 6 7 '5 feet nearly
This compares favourably with the actual result of 69*9 feet.
The quantities required for the use of the formula would all be
known at a very early stage of a design and a close approxima
tion to the height H can thus very readily be obtained. A
formula such as this is useful as a check on the result of the
calculation for the longitudinal BM.
We may also obtain an approximate formula in the same
manner as was done for the transverse BM on p. in. Using
a similar system of notation, we may say
Moment of inertia of L.W.P. about a trans
T * v
verse axis through the centre of flotation f
n f being a coefficient of a similar nature to n used on p. 107.
Volume of displacement = xLxBxD
tt X L 3 X B
"' ~
where b is a coefficient obtained from the coefficients n' and k.
Sir William White, in the " Manual of Naval Architecture," says,
with reference to the value of , that "the value 0075 ma y be
used as a rough approximation in most cases ; but there are
many exceptions to its use." If this approximation be applied
to the example we have worked, the mean moulded draught
being 5*8 feet
The value of H = 65 feet
152 Theoretical Naval Architecture.
This formula shows very clearly that the length of a ship is
more effective than the draught in determining the value of the
longitudinal BM in any given case. For vessels which have an
unusual proportion of length to draught, the values of the longi
tudinal BM found by using this formula will not be trustworthy.
To estimate the Displacement of a Vessel when
floating out of the Designed Trim. The following
method is found useful when it is not desired to actually
calculate the displacement from the drawings, and a close
approximation is sufficiently accurate. Take a ship floating
parallel to her designed L.W.L. ; we can at once determine
the displacement when floating at such a waterline from the
curve of displacement (see p. 25). If now a weight already
on board is shifted aft, say, the ship will change trim, and she
will trim more by the stern than designed. The new water
plane must pass through the centre of gravity of the original
waterplane, or, as we have termed it, the centre of flotation, and
FIG. 63.
the displacement at this new waterline will be the same as at
the original waterline. Now, when taking the draught of water
a vessel is actually floating at, we take the figures set up at or
near the forward and after perpendiculars. These draughts,
if not set up at the perpendiculars, can be transferred to the
perpendiculars by a simple calculation. The draughts thus
obtained are added together and divided by two, giving us
the mean draught. Now run a line parallel to the designed
waterline at this mean draught, as in Fig. 63, where WL
represents the actual waterline, and wl the line just drawn.
It will not be true that the displacement of the ship is the same
as that given by the waterline wl. Let F be the centre of
flotation of the waterline wt, and draw WL' through F parallel
to WL. Then the actual displacement will be that up to WL',
which is nearly the same as that up to wl, with the displacement
Longitudinal Metacentre, Longitudinal J3M, etc. 153
of the layer WW'L'L added. The displacement up to wl is
found at once from the curve of displacement. Let T be the
tons per inch at //, and therefore very nearly the tons per inch
at W'L' and WL. SF, the distance the centre of flotation of
the waterplane wl is abaft the middle of length, is supposed
known, and equals d inches, say. Now, the angle between wl
and WL is given by
tan0 =
length of ship
_ amount out of normal trim
length of ship
But if x is the thickness of layer in inches between W'L' and
WL, we also have in the triangle SFH
tan 6 = j very nearly (for small angles tan = sin
very nearly)
and accordingly x may be determined. This, multiplied by
the tons per inch T, will give the displacement of the layer. 1
The following example will illustrate the above :
Example. A vessel floats at a draught of 16' 5^" forward, 23' ij" aft,
the normal trim being 2 feet by the stern. At a draught of 19' 9$", her
displacement, measured from the curve of displacement, is 5380 tons, the
tons per inch is 31*1 tons, and the centre of flotation is 12 '9 feet abaft
amidships. Estimate the ship's displacement.
The difference in draught is 23' i" 1 6' 5J" = 6' 8", or 4' 8" out of
trim. The distance between the draughtmarks is 335 feet, and we
therefore have for the thickness of the layer
12 x 129 X * = 215 inches
The displacement of the layer is therefore
215 X 311 = 67 tons
The displacement is therefore
5380 + 67 = 5447 tons nearly
Change of Trim due to Longitudinal Shift of
Weights already on Board. We have seen that change
1 This may be reduced to a formula, set as an example in Appendix A.
T X y
No. 2, viz. extra displacement for I foot extra trim = 12 j , y being
centre of flotation abaft amidships in feet.
154
Theoretical Naval Architecture.
of trim is the sum of the change of draughts forwaid and aft, and
that change of trim can be caused by the shift of weights on
board in a foreandaft direction. We have here an analogous
case to the inclining experiment in which heeling is caused by
shifting weights in a transverse direction. In Fig. 64, let w be
FIG. 64.
a weight on the deck when the vessel is floating at the water
line WL, G being the position of the centre of gravity. Now
suppose the weight w to be shifted forward a distance of d feet.
G will, in consequence of this, move forward parallel to the line
joining the original and final positions of a/, and if W be the
displacement of the ship in tons, G will move to G' such that
w X d
GG' =
W
Now, under these circumstances, the condition of equilibrium
is not fulfilled if the waterline remains the same, viz. that the
centre of gravity and the centre of buoyancy must be in the
same vertical line, because G has shifted to G'. The ship
must therefore adjust herself till the centre of gravity and the
centre of buoyancy are in the same vertical line, when she
will float at a new waterline, W'L', the new centre of buoyancy
being B'. The original vertical through G and B meets the
new vertical through G' and B' in the point M, and this point
will be the longitudinal metacentre, supposing the change of
trim to be small, and GM will be the longitudinal metacentric
height. Draw W'C parallel to the original waterline WL.
Longitudinal Metacentre, Longitudinal BM, etc. 155
meeting the forward perpendicular in C. Then, since CL =
W'W, the change of trim WW'+ LL'= CL' = #, say. The
angle of inclination of W'L' to WL is the same as the angle
between W'L' and W'C = 6, say, and
CL' x
tan = ;  r = T
length L
But we also have
therefore, equating these two values for tan 6, we have
x _GG'
L GM
w X d .
"" W X GM
using the value obtained above for GG' ; or
*, the change of trim due to the "j ^
moment of transference of the > = \y x r* vf X L feet
weight a/ through the distance </, j
or
. . .., 12 X w X d?X L
The change of trim in inches =  w v GM
and the moment to change trim i inch is
W x GM r
w X d = = foottons
1 2 X \i
To determine this expression, we must know the vertical
position of the centre of gravity and the position of the longi
tudinal metacentre. The vertical position of the centre of
gravity will be estimated in a design when dealing with the
metacentric height necessary, and the distance between
the centre of buoyancy and the centre of gravity is then sub
tracted from the value of the longitudinal BM found by one of
the methods already explained. The distance BG is, however,
small compared with either of the distances BM or GM and
any small error in estimating the position of the centre of
gravity cannot appreciably affect the value of the moment to
change trim one inch. In many ships BM approximately
156 Theoretical Naval Architecture.
equals the length of the ship, and therefore GM also ; we may
therefore say that in such ships the moment to change trim
i inch = pa the displacement in tons. For ships that are long
in proportion to the draught, the moment to change trim i inch
is greater than would be given by this approximate rule.
In the ship for which the value of the longitudinal BM was
calculated on p. 148, the centre of buoyancy was 2\ feet below
the L.W.L., the centre of gravity was estimated at i^ feet
below the L.W.L. ; and the length between perpendiculars was
75 f eet.
/. GM = 699  i
= 689 feet
and the moment to change trim i inch =
12 x 75
= 7 'oi foottons
the draughts being taken at the perpendiculars.
Example. A vessel 300 feet long and 2200 tons displacement has a
longitudinal metacentric height of 490 feet. Find the change of trim
caused by moving a weight of 5 tons already on board through a distance
of 200 feet from forward to aft.
Here the moment to change trim I inch is
'I^f = 300 foottons near l y
The moment aft due to the shift of the weight is
5 X 200 = looo foottons
and consequently the change of trim aft is
*ffl = 3$ inches
Approximate Formula for the Moment to change Trim i inch.
Assuming Normand's approximate formula for the height
of the longitudinal metacentre above the centre of buoyancy
given on p. 151
A 2 X L
H = 00735 g^
we may construct an approximate formula for the moment to
change trim i inch as follows.
We have seen that the moment to change trim i inch is
W x GM
12 x L
Longitudinal Metacentre, Longitudinal BM, etc. 157
V
We can write W = and assume that, for our purpose,
OJ
BM = GM = 0
Substituting this in the above formula, we have
Moment to change  _ V f A 2 x L \
trim i inch ) 3 5Xi2XLV '735 B x v )
A 2
or 0*000175^5
For further approximations, see Example 18, p. 173.
Applying this to the case worked out in detail on p. 148
Area of L.W.P. = A = 773*5 square feet
Breadth = B = 1442 feet
so that the moment to change trim i inch approximately
should equal
(77VO 2
o*oooi75 v /J J/ = 7*26 foottons
5 1442
the exact value, as calculated on p. 156, being 7*01 foottons.
It is generally sufficiently accurate to assume that onehalf
the change of trim is forward, and the other half is aft. In the
example on p. 156, if the ship floated at a draught of 12' 3"
forward and 14' 9" aft, the new draught forward would be
1 2' 3" if"= 1 2' 4"
and the new draught aft would be
14' 9" + '?' = H' iof
Referring, however, to Fig. 64, it will be seen that when,
as is usually the case, the centre of flotation is not at the middle
of the length, WW' is not equal to LL', so that, strictly speak
ing, the total change of trim should not be divided by 2, and
onehalf taken forward and the other half aft. Consider the
triangles FWW, FLL'; these triangles are similar to one
158 Theoretical Naval Architecture.
another, and the corresponding sides are proportional, so
that
WW ; LL'
~ == LF
and both these triangles are similar to the triangle W'CL'.
Consequently
WW' _ LU _ CU _ change of trim
WF ~ LF ~ W'C " length
WF
/. WW' = .  T x change of trim
length
T F
and LL' = x change of trim
that is to say, the proportion of the change of trim either aft or
forward, is the proportion the length of the vessel abaft or
forward of the centre of flotation bears to the length of the
vessel. Where the change of trim is small, this makes no
appreciable difference in the result, but there is a difference
when large changes of trim are under consideration.
For example, in the case worked out on p. 156, suppose
a weight of 50 tons is moved through 100 feet from forward to
aft ; the change of trim caused would be
i6 inches
The centre of flotation was 1 2 feet abaft the middle of length.
The portion of the length abaft the centre of flotation is there
fore Ml f tne length. The increase of draught aft is there
fore
138 v 8.0. ,2. i nr Vips
300* 3 13 1Ilcilc:s
and the decrease of draught forward is
162 y _P_ _ n i nr hpi
300 * 3 9 "H"Gb
instead of 8 inches both forward and aft. The draught
forward is therefore
1 2' 3" 9"= n' 6"
and the draught aft
M' 9" + 7f = 15' 4f"
It will be noticed that the mean draught is not the same as
Longitudinal Metacentre, Longitudinal BM, etc. 159
before the shifting, but twothirds of an inch less, while the
displacement remains the same. This is due to the fact that,
as the ship increases her draught aft and decreases it forward, a
fuller portion of the ship goes into the water and a finer portion
comes out.
Effect on the Trim of a Ship due to adding a
Weight of Moderate Amount. If we wish to place a
weight on board a ship so that the vessel will not change trim,
we must place it so that the upward force of the added buoyancy
will act in the same line as the downward force of the added
weight. Take a ship floating at a certain waterline, and
imagine her to sink down a small amount, so that the new
waterplane is parallel to the original waterplane. The added
buoyancy is formed of a layer of parallel thickness, and having
very nearly the shape of the original waterplane. The upward
force of this added buoyancy will act through the centre of
gravity of the layer, which will be very nearly vertically over
the centre of gravity of the original waterplane, or, as we have
termed it, the centre of flotation. We therefore see that to
place a weight of moderate amount on a ship so that no
change of trim takes place, we must place it vertically over or
under the centre of flotation. The ship will then sink to a new
waterline parallel to the original waterline, and the distance
she will sink is known at once, if we know the tons per inch
at the original waterline. Thus a ship is floating at a draught
of 13 feet forward and 15 feet aft, and the tons per inch immer
sion is 20 tons. If a weight of 55 tons be placed over or under
the centre of flotation, she will sink ff inches, or 2f inches,
and the new draught will be 13' 2f" forward and 15' 2f" aft.
It will be noticed that we have made two assumptions, both
of which are rendered admissible by considering that the weight
is of moderate amount. First, that the tons per inch does not
change appreciably as the draught increases, and this is, for all
practical purposes, the case in ordinary ships. Second, that the
centre of gravity of the parallel layer of added buoyancy is in
the same section as the centre of flotation. This latter assump
tion may be taken as true for small changes in draught caused
by the addition of weights of moderate amount ; but for large
160 Theoretical Naval Architecture.
changes it will not be reasonable, because the centres of gravity
of the waterplanes are not all in the same section, but vary for
each waterplane. As a rule, waterplanes are fuller aft than
forward near the L.W.P., and this more so as the draught
increases ; and so, if we draw on the profile of the sheer drawing
a curve through the centres of gravity of waterplanes parallel to
the L.W.P., we should obtain a curve which slopes somewhat
aft as the draught increases. We shall discuss further the
methods which have to be adopted when the weights added
are too large for the above assumptions to be accepted.
We see, therefore, that if we place a weight of moderate
amount on board a ship at any other place than over the centre
of flotation, she will not sink in the water to a waterline
parallel to the original waterline, but she will change trim as
well as sink bodily in the water. The change of trim will be
forward or aft according as the weight is placed forward or
aft of the centre of flotation.
In determining the new draught of water, we proceed in
two steps :
1. Imagine the weight placed over the centre of flotation,
and determine the consequent sinkage.
2. Then imagine the weight shifted either forward or aft to
the assigned position. This shift will produce a certain moment
forward or aft, as the case may be, equal to the weight multiplied
by its longitudinal distance from the centre of flotation. This
moment divided by the moment to change trim i inch as cal
culated for the original waterplane will give the change of trim.
The steps will be best illustrated by the following example :
A vessel is floating at a draught of 12' 3" forward and 14' 6" aft. The
tons per inch immersion is 2O ; length, 300 feet ; centre of flotation, 12 feet
abaft the middle of length ; moment to change trim I inch, 300 foottons.
A weight of 30 tons is placed 20 feet from the forward end of the ship.
What will be the new draught of water ?
The first step is to see the sinkage caused by placing the weight over
the centre of flotation. This sinkage is i inches, and the draughts would
then be
12' 4$" forward, 14' 7J" aft
Now, the shift from the centre of flotation to the given position is 142
feet, so that the moment forward is 30 X 142 foottons, and the change
of trim by the bow is
22 4_, or 14 \ inches nearly
300
Longitudinal Metacentre, Longitudinal BM, etc. 161
This has to be divided up in the ratio of 138 : 162, because the centre
of flotation is 12 feet abaft the middle of length. We therefore have
Increase of draught forward g X 14!" = 7f say
Decrease of draught aft $j X 14^" = 6^" say
The final draughts will therefore be
Forward, 12' 4^" + 7?" = 13' oi"
Aft, 14' 7*"  64" = M' I"
Effect on the Trim of a Ship due to adding a
Weight of Considerable Amount. In this case the
assumptions made in the previous investigation will no longer
hold, and we must allow for the following :
1. Variation of the tons per inch immersion as the ship
sinks deeper in the water.
2. The centre of flotation does not remain in the same
transverse section.
3. The addition of a large weight will alter the position
of G, the centre of gravity of the ship.
4. The different form of the volume of displacement will
alter the position of B, the centre of buoyancy of the ship, and
also the value of BM.
5. Items 3 and 4 will alter the value of the moment to
change trim i inch.
As regards i, we can obtain first an approximation to the
sinkage by dividing the added weight by the tons per inch
immersion at the original waterline. The curve of tons per
inch immersion will give the tons per inch at this new draught.
The mean between this latter value and the original tons per
inch, divided into the added weight, will give a very close
approximation to the increased draught. Thus, a vessel floats at
a constant draught of 22' 2", the tons per inch immersion
being 44*5. It. is requiredi to find the draught after adding a
weight of 750 tons. The first approximation to the increase of
draught is ^ =17 inches nearly. At a draught of 23' 7"
44'5
it is found that the tons per inch immersion is 457. The
mean tons per inch is therefore (44*5 + 45*7) = 45' 1 ) an d
the increase in draught is therefore m  = 1663, or J 6f inches
M
1 62 Theoretical Naval Architecture.
nearly. This assumes that the ship sinks to a waterplane
parallel to the first waterplane. In order that this can be the
case, the weight must have been placed in the same transverse
section as the centre of gravity of the layer of displacement
between the two waterplanes. We know that the weight and
buoyancy of the ship must act in the same vertical line, and
therefore, for the vessel to sink down without change of trim,
the added weight must act in the same vertical line as the
added buoyancy. We can approximate very closely to the
centre of gravity of the layer as follows : Find the centre of
flotation of the original W.P. and that of the parallel W.P.
to which the vessel is supposed to sink. Put these points on
the profile drawing at the respective waterlines. Draw a line
joining them, and bisect this line. Then this point will be
a very close approximation to the centre of gravity of the layer.
A weight of 750 tons placed as above, with its centre of gravity
in the transverse section containing 'this point, will cause the
ship to take up a new draught of 23' 6f " with no change of trim.
We can very readily find the new position of G, the centre
of gravity of the ship due to the addition of the weight. Thus,
suppose the weight of 750 tons in the above example is placed
with its centre of gravity 16 feet below the C.G. of the ship;
then, supposing the displacement before adding the weight to
be 9500 tons, we have
(750 x 16
Lowering of G =
10250
= 1*17 feet
We also have to take account of 4. In the case we have
taken, the new C.B. below the original waterline was 97 feet,
as against io'5 feet in the original condition, or a rise of 0*8
foot.
For the new waterplane we have a different longitudinal
BM, and, knowing the new position of B and of G, we can deter
mine the new longitudinal metacentric height. From this we
can obtain the new moment to change trim i inch^ using, of
course, the new displacement. In the above case this works
out to 950 foottons.
Longitudinal Metacentre, Longitudinal BM, etc. 163
Now we must suppose that the weight is shifted from the
assumed position in the same vertical line as the centre of
gravity of the layer to its given position, and this distance must
be found. The weight multiplied by the longitudinal shift will
give the moment changing the trim either aft or forward, as the
case may be. Suppose, in the above case, this distance is 50 feet
forward. Then the moment changing trim by the bow is
750 X 50 = 37,500 foottons
and the approximate change of trim is
37,5 T 95 = 39* inches
This change of trim has to be divided up in the ordinary
way for the change of draught aft and forward. In this case
we have
Increase of draught forward = ~\ X 39^ = 21^ inches say
Decrease of draught aft = Jff X 39^= 18 inches say
We therefore have for our new draughts
Draught aft, 22' 2" + i6f"  18" =22' of"
Draught forward, 22' 2" + i6f" + 21*" = 25' 4"
For all ordinary purposes this would be sufficiently accu
rate ; but it is evidently still an approximation, because we do
not take account of the new GM for the final waterline, and
the consequent new moment to change trim i inch. These can
be calculated if desired, and corrections made where necessary.
To determine the Position of a Weight on Board
a Ship such that the Draught aft shall remain
constant whether the Weight is or is not on Board. 1
Take a ship floating at the waterline WL, as in Fig. 65. If
a weight w be placed with its centre of gravity in the transverse
section that contains the centre of flotation, the vessel will very
nearly sink to a parallel waterline W'L'. 2 This, however, is
not what is required, because the draught aft is the distance
WW greater than it should be. The weight will have to be
1 See also Examples 25, 26 in Appendix.
* Strictly speaking, the weight should be placed with its centre of
gravity in the transverse section that contains the centre of gravity of the
zone between the waterlines WL and W'L'.
164 Theoretical Naval Architecture.
moved forward sufficient to cause a change of trim forward of
WW f LL', and then the draught aft will be the same as it
originally was, and the draught forward will increase by the
amount WW f LL'. This will be more clearly seen, perhaps,
by working the following example :
It is desired that the draught of water aft in a steamship
(particulars given below) shall be constant, whether the coals
FIG. 65.
are in or out of the ship. Find the approximate position of
the centre of gravity of the coals in order that the desired
condition may be fulfilled: Length of ship, 205 feet; displace
ment, 522 tons (no coals on board) ; centre of flotation from
after perpendicular, 104*3 f eet ; longitudinal BM, 664 feet;
longitudinal GM, 661*5 feet; tons per inch, 11*4; weight of
coals, 57 tons.
From the particulars given, we find that
Moment to change ) 661*5 x 522
. , } = = 140 foottons
trim i inch j 12 X 205
The bodily sinkage, supposing the coals placed with the centre
of gravity in the transverse section containing the centre of
flotation, will be ** = 5 inches. Therefore the coals must
11*4
be shifted forward from this position through such a distance
that a change of trim of 10 inches forward is produced.
Accordingly, a forward moment of
140 x 10 = 1400 foottons
is required, and the distance forward of the centre of flotation
the coals require shifting is
= 246 feet
Longitudinal Metacentre, Longitudinal BM, etc 165
Therefore, if the coals are placed
104*3 + 2 4'6 = 1289 feet
forward of the after perpendicular, the draught aft will remain
very approximately the same as before.
Change of Trim caused by a Compartment being
open to the Sea. The principles involved in dealing with
a problem of this character will be best understood by working
out the following example :
A rectangularshaped lighter, 100 feet long, 40 feet broad,
10 feet deep, floating in salt water at 3 feet level draught, has
a collision bulkhead 6 feet from the forward end. If the side
is broached before this bulkhead below water, what would be
the trim in the damaged condition ?
Let ABCD, Fig. 66, be the elevation of the lighter, with a
PIG. 66.
collision bulkhead 6 feet from the forward end, and floating
at the level waterline WL. It is well to do this problem
in two stages
1. Determine the amount of mean sinkage due to the loss
of buoyancy.
2. Determine the change of trim caused.
i. The lighter, due to the damage, loses an amount of
buoyancy which is represented by the shaded part GB, and if
we assume that she sinks down parallel, she will settle down at
a waterline wl such that volume wG = volume GB. This
will determine the distance x between wl and VVL.
1 66 Theoretical Naval Architecture.
For the volume wG = wH X 40 feet X x
and the volume GB = GL x 40 feet X 3 feet
40 x 6 x 3 18 f
.'. x =  if feet
94 X 40
= 2\ inches nearly
2. We now deal with the change of trim caused.
The volume of displacement = 100 x 4 X 3 cubic feet
The weight of the lighter = I0 X 40 X 3 = 2^0 tons
oo
and this weight acts down through G, the centre of gravity,
which is at 50 feet from either end.
But we have lost the buoyancy due to the part forward of
bulkhead EF, and the centre of buoyancy has now shifted
back to B' such that the distance of B' from the after end is
47 feet. Therefore we have W, the weight of lighter, acting
down through G, and W, the upward force of buoyancy, acting
through B'. These form a couple of magnitude
W x 3 feet = ^P X 3 = ^^ foottons
tending to trim the ship forward.
To find the amount of this trim, we must find the moment
to change trim i inch
_WxGM
12 x L
using the ordinary notation.
Now, GM very nearly equals BM ;
9400
.". moment to change trim i inch = 7 x BM
12 X IOO
= X BM
where I = the moment of inertia of the intact waterplane about
a transverse axis through its centre of gravity ;
and V = volume of displacement in cubic feet.
Longitudinal Metacentre, Longitudinal BM, etc. 167
I = ir(94 X 4) X (94)'
V = 12000
I440OO
2 X 40 X (94)*
and moment to alter trim i inch = 
7 X 144000
= 66 foottons nearly
.*. the change of trim = ^p 4 66
= 15^ inches
The new waterline W'L' will pass through the centre
of gravity of the waterline wl at K, and the change of trim
aft and forward must be in the ratio 47 : 53 ; or
Decrease of draught aft = T V^ x 15^ = 1\ inches
Increase of draught forward == f^ X 15^ = 8^ inches
therefore the new draught aft is given by
3' o" + 2j"  7 f = 2' 7"
and the new draught forward by
3' o" + 2 i" + 8J" = 3' loi"
The correctness of this result may be seen by finding the
displacement and position of the C.B. of the new volume of
displacement. The displacement will be found equal to the
original displacement, and the C.B. will be found to be 50 feet
from the afterend or the same as the C.G.
(For a more difficult example of a similar nature, see No.
34, Appendix A.)
If the sums of the columns 4, 6, and 8 in the table on
p. 148 are called Sj, S 2 , S 3 , then Area = S, X f . h, and the
distance of the C.G. of waterplane from No. 6, viz. :
S /z 3
s = ?X^;I n = S 3 X2X
ch 3
I the moment of inertia required.
168
Theoretical Naval Architecture.
This method saves some work as compared with the above
and is used in Brown's displacement sheet given in the Ap
pendix. In the above case
7iT (56'S6) 2 i
I = 2 . ' 959 14  v ^
3 L 7J 16342 J
This worked out by the aid of the 4 fig. logs, gives 224,300.
To find the Longitudinal Position of the C.G. of a
Ship. If a ship is floating at the trim assumed for the ordinary
calculations, this is a simple matter, as the C.G. must be in
the same vertical line as the C.B., and the longitudinal position
of the C.B. is readily found for the draught at which the ship
is floating. If, however, the ship is floating out of her normal
trim, the following gives a close approximation to the position
of the C.G.
Suppose the ship is trimming by the stern at the waterline
WL, as in Fig. 66A. The waterline cutting off the same
displacement is not wl at the same mean draught as WL, but
w'l' passing through F, the centre of flotation. The excess
displacement over that corresponding to the mean draught is
i2X.yxTx0, where T is the tons per inch, y the C.F.
abaft amidships in feet, 6 is the angle between WL and wl
(i.e. change of trim 4 length). The C.B. of the displacement
corresponding to w'l' is at B , and can readily be determined.
The ship, in trimming to the waterline WL, may be said to
Longitudinal Metacentre^ Longitudinal BM, etc. 169
pivot about a transverse axis through F, and the volume F/'L
shifts to Ww'F. Then it can readily be shown that the stern
ward shift of the C.B. from B to B is BM L X 0, BM L being
the longitudinal BM corresponding to w'l' or WL. The
C.G. of the ship must be in the line BM L perpendicular to
WL, and therefore G abaft midlength = B abaft midlength
(BGX0).
If the ship trims by the bow, the C.B. shifts forward
BM L x 0, and the C.G. is B abaft midlength + (BG X 0).
Change of Trim due to passing from Salt to River
Water, or viceversa, If the C.F. is vertically over the
C.B. there will be no change of trim. If W is weight, and
V and V are the volumes of displacement in salt water and
river water (say 35 and 356 cubic feet to the ton respectively),
then V = W x 35, V = W x 35'6. Let V'V = v.
The vessel floating at the line WL in sea water B and G
must be in the same vertical. Supposing the ship to sink
down parallel to W'L' in river water, the shift aft of the C.B. is
, where b is the fore and aft separation of the C.B. and
v . .
C.F., and the moment changing trim = W X y , X b. The
following example will illustrate the above.
Example. A vessel with rectangular sections is 300 feet long, 30 feet
broad, and floats in salt water at a draught of 15 feet forward and 20 feet aft.
C.G. in WL. Determine the draughts forward and aft on going into
river water 63 Ibs. to cubic foot.
W
Bodily sinkage =r = 3' 34 inches.
Volume of layer = 300 x 30 x 334 X T ' 5 = 2505 cubic feet.
C.B. and .*. C.G. abaft amidships =7*15 feet.
C.F. from amidships = nil.
Moment to change trim one inch = 525 feet tons.
W = 4500 tons V = 157,500 v = 2505.
Therefore shift (forward in this case) of C.B. = x 7*15 = O'l I ft.
The C.G. and the new C.B. are therefore o'li feet apart, and the
moment to change trim aft is 4500 X o'li feet tons, and the change of
trim is I inch, say.
The draughts are therefore 15' 3 "34" 0*5" =15' 2'84".
,, ,, ,, 20' 334" +05" = 20' 384".
17 Theoretical Naval Architecture.
Draught of a Vessel when Launched. It is fre
quently necessary to make a close approximation to the
draught forward and aft of a vessel on the occasion of launch
ing, and in addition to the ordinary hydrostatic curves given
in Fig. 153 it is necessary to obtain the weight of the vessel
on the stocks and the position of the C.G. of this weight, both
in a longitudinal and a vertical direction. The weight will
enable the mean draught to be fixed, taking into account the
density of the water. At this draught the position of the
longitudinal metacentre is known, from which the longitudinal
GM can be found, and then the moment to change trim i inch.
The longitudinal centre of buoyancy at the assumed draught
can be found readily, and the moment changing trim is deter
mined by multiplying the weight and the longitudinal distance
between the centre of buoyancy and the centre of gravity.
The following example will illustrate the methods to be
adopted
A boxshaped vessel, 400 feet by 70 feet, floating when at
designed draught at 22 feet forward and 24 feet aft, weighs
before launching 6400 tons, and the position of the centre of
gravity is i o feet abaft amidships and 3 feet below L. W.L.
What will be her draught whe?i launched into salt water ?
The mean draught is 8 feet, and assume she floats parallel
to the L.W.L., 7 feet forward and 9 feet aft. At this waterline
the C.B. is readily calculated to be 83 feet abaft amidships
and 19*0 feet below L.W.L. The longitudinal BM at this
assumed waterline works out to 1,666 feet, and the longitudinal
GM 1650 feet, since BG is 16 feet. The moment to change
trim i inch is 2200 foot tons. The horizontal separation of
the C.G. and the C.B. is 10 8*3 =17 feet, so that the
change of trim is  2200 = 5 incnes aft  In tnis
case, seeing that the centre of flotation is amidships, this
5 inches is divided equally forward and aft, so that the draught
when launched is 6 ft. 9^ in. forward and 9 ft. 2^ in. aft.
The principal difficulty in such an estimate is the deter
mination of the longitudinal C.G.
Longitudinal Metacentre, Longitudinal BM, etc. 171
Information for use when Docking in a Floating
Dock. When a ship is to be docked in a floating dock,
especially if the weight is close to the lifting capacity of the
dock, it is necessary to place the ship in the dock so that its
C.G. is at the centre of length of the dock in order that when
lifted the dock shall be on an even keel.
Information in the following form is now provided to
H.M. ships to enable the position of the C.G. to be closely
approximated to knowing the draughts forward and aft. This
information can be readily calculated from the sheer drawing
by using the principles of the present chapter.
Mean draught
between
draught marks.
Corresponding
displacement at
normal trim.
Longitudinal
position of centre
of buoyancy
relative
to bulkhead 132.
Movement of the C.B.
for every foot change
of trim from normal
trim by bow or stern.
Tons
per
inch.
ftt in.
tons.
feet.
feet.
tons.
31 6
25,900
15*7 forward
129
*3
28 6
22,930
I7'i
142
82
256
2O,OOO
190 ,,
F 5 8
81
Thus, suppose the above ship is drawing 28 ft. forward
and 31 ft. 6 in. aft, the C.B. at the mean draught of 29 ft. 9 in.
even keel is 16*4 ft. forward of 132, and this will move aft
35 x 1*36 = 475 ft. for the 3 ft. 6 in. trim by the stern.
The C.B. (and therefore the C.G. very nearly) is therefore
1 64 4*75 = 11*65 ft forward of 132 station, and this point
should as nearly as possible be placed at the centre of length
of the dock.
Examples. (i) In the above ship if the draught forward is 25 ft. 6 in.,
and the draught aft 32 ft., estimate the position the ship should be placed
relative to the dock.
Ans. 7*8' forward of 132 should be well with dock centre,
(ii) In the above ship if the draught forward is 30 ft., and aft 25 ft.,
estimate the position the ship should be placed relative to the dock.
Ans. 25' forward of 132 should be well with dock centre.
172 Ttieoretical Naval Architecture.
EXAMPLES TO CHAPTER IV.
I . A ship is floating at a draught of 20 feet forward and 22 feet aft, when
the following weights are placed on board in the positions named :
Weight Distance from C.G. of
in tons. waterplane in feet.
'}**
}
What will be the new draught forward and aft, the moment to change
trim I inch being 800 foottons, and the tons per inch = 35 ?
Ans. 20' 5 f " forward, 22' 3" aft.
2. A vessel 300 feet long, designed to float with a trim of 3 feet by
the stern, owing to consumption of coal and stores, floats at a draught of
9' 3" forward, and 14' 3" aft. The load displacement at a mean draught of
13' 6" is 2140 tons ; tons per inch, i8 ; centre of flotation, 12$ feet abaft the
middle of length. Approximate as closely as you can to the displacement.
Ans. 1775 tons.
3. A vessel is 300 feet long and 36 feet beam. Approximate to the
moment to change trim I inch, the coefficient of fineness of the L.W.P.
being 0*75.
Ans. 319 foottons.
4. A lightdraught sternwheel steamer is very approximately of the form
of a rectangular box of 1 20 feet length and 20 feet breadth. When fully
laden, the draught is 18 inches, and the centre of gravity of vessel and
lading is 8 feet above the waterline. Find the transverse and longitudinal
metacentric heights, and also the moment to change trim one inch.
Ans. 1347 feet, 79 1^ feet ; 56^ foottons.
5. A vessel is floating at a draught of 12' 3" forward and 14' 6" aft.
The tons per inch immersion is 20 j length, 30x3 feat ; centre of flotation,
12 feet abaft amidships; moment to change trim I inch, 300 foottons.
Where should a weight of 60 tons be placed on this vessel to bring her to
an even keel.
Ans. 123 feet forward of amidships.
6. What weight placed 13 feet forward of amidships will have the same
effect on the trim of a vessel as a weight of 5 tons placed 10 feet abaft the
forward end, the length of the ship being 300 feet, and the centre of
flotation 12 feet abaft amidships.
Ans. 30 '4 tons.
7. A right circular pontoon 50 feet long and 16 feet in diameter is just
half immersed on an even keel. The centre of gravity is 4 feet above the
bottom. Calculate and state in degrees the transverse heel that would be
produced by shifting 10 tons 3 feet across the vessel. State, in inches, the
change of trim produced by shifting 10 tons longitudinally through 20 feet.
Ans. 3 degrees nearly ; 25 inches nearly.
8. Show why it is that many ships floating on an even keel will increase
the draught forward, and decrease the draught aft, or, as it is termed, go
down by the head, if a weight is placed at the middle of the length.
9. Show that for vessels having the ratio of the length to the draught
about 13, the longitudinal B.M. is approximately equal to the length.
Why should a shallow draught river steamer have a longitudinal B.M.
much greater than the length ? What type of vessel would have a longitu
dinal B.M. less than the length ?
Longitudinal Metacentre, Longitudinal BM> etc. 173
10. Find the moment to change trim I inch of a vessel 400 feet long,
having given the following particulars : Longitudinal metacentre above
centre of buoyancy, 446 feet ; distance between centre of gravity and centre
of buoyancy, 14 feet ; displacement, 15,000 tons.
Ans. 1350 foottons.
11. The moment of inertia of a waterplane of 22,500 square feet
about a transverse axis 20 feet forward of the centre of flotation, is found
to be 254,000,000 in footunits. The displacement of the vessel being
14,000 tons, determine the distance between the centre of buoyancy and
the longitudinal metacentre.
Ans. 500 feet.
12. In the preceding question, if the length of the ship is 405 feet, and
the distance between the centre of buoyancy and the centre of gravity is 13
feet, determine the change of trim caused by the longitudinal transfer of
150 tons through 50 feet.
Ans. 5 1 inches nearly.
13. A water plane has an area of 13,200 square feet, and its moment of
inertia about a transverse axis 14^ feet forward of its centre of gravity
works out to 84,539,575 in footunits. The vessel is 350 feet long, and
has a displacement to the above waterline of 5600 tons. Determine the
moment to change trim I inch, the distance between the centre of gravity
and the centre of buoyancy being estimated at 8 feet
Ans. 546 foottons.
14. The semiordinates of a waterplane of a ship 20 feet apart are as
follows: 04, 75, 145, 2i'o, 266, 309, 340, 360, 370, 373, 373,
37'3. 37'3> 37'2, 37'i> 36'8, 35'8, 33'4> 288, 21 7, 115 feet respectively.
The after appendage, whole area 214 square feet, has its centre of gravity
6'2 feet abaft the last ordinate. Calculate
1 i ) Area of the waterplane.
(2) Position of C.G. of waterplane.
(3) Transverse B.M.
(4) Longitudinal B.M.
(Volume of displacement up to the waterplane 525,304 cubic feet.)
Ans. (i) 24,015 square feet; (2) 18*2 feet aoaft middle ordinate;
(3) 1716 feet; (4) 4476 feet.
15. The semiordinates of the L.W.P. of a vessel 15! feet apart are,
commencing from forward, o'l, 2'5, 53, 8'i, io'8, 131, 15*0, 16*4, 176,
183, 185, 185, 184, 181, 175, 166, 153, 133, 108, 76, 38 feet
respectively. Abaft the last ordinate there is a portion of the waterplane,
the halfarea being 27 square feet, having its centre of gravity 4 feet abaft
the last ordinate. Calculate the distance of the longitudinal metacentre
above the centre of buoyancy, the displacement being 2206 tons.
Ans. 534 feet.
16. State the conditions that must hold in order that a vessel shall not
change trim in passing from river water to salt water.
17. A log of fir, specific gravity o'5, is 12 feet long, and the section is
2 feet square. What is its longitudinal metacentric height when floating in
stable equilibrium ?
Ans. i6'5 feet nearly.
1 8. Using the approximate formula for the moment to change trim i
inch given on p. 157, show that this moment will be very nearly given by
30 . =, where T is the tons per inch immersion, and B is the breadth.
Show also that in ships of ordinary form, the moment to change trim
i inch approximately equals ^^ . L 2 B.
CHAPTER V.
STATICAL STABILITY, CURVES OF STABILITY, CALCU
LATIONS FOR CURVES OF STABILITY, INTEGRATOR,
DYNAMICAL STABILITY.
Statical Stability at Large Angles of Inclination.
Atwood's Formula. We have up to the present only dealt
with the stability of a ship at small angles of inclination, and
within these limits we can determine what the statical stability is
by using the metacentric method as explained on p. 98. We
must now, however, investigate how the statical stability of a
ship can be determined for large angles of inclination, because
in service it is certain that she will be heeled over to much
larger angles than 10 to 15, which are the limits beyond which
we cannot employ the metacentric method.
Let Fig. 67 represent the crosssection of a ship inclined
to a large angle 6. WL is the position on the ship of the
original waterline, and B the original position of the centre of
buoyancy. In the inclined position she floats at the waterline
W'L', which intersects WL in the point S, which for large angles
will not usually be in the middle line of the ship. The volume
SWW' is termed, as before, the " emerged wedge" and the volume
SLL' the " immersed wedge" and g; g 1 are the positions of the
centres of gravity of the emerged and immersed wedges respec
tively. The volume of displacement remains the same, and
consequently these wedges are equal in volume. Let this
volume be denoted by v. The centre of buoyancy of the
vessel when floating at the waterline WL' is at B', and the
upward support of the buoyancy acts through B'; the downward
force of the weight acts through G, the centre of gravity of the
ship. Draw GZ and BR perpendicular to the vertical through
B', and gh,gti perpendicular to the new waterline W'L'. Then
Statical Stability, Curves of Stability, etc. 175
the moment of the couple tending to right the ship is W x GZ,
or, as we term it, the moment of statical stability. Now
GZ = BR  BP
= BR  BG sin
so that the moment of statical stability at the angle 6 is
W(BR BG.sinfl)
The length BR is the only term in this expression that we
do not know, and it is obtained in the following manner. The
new volume of displacement W'AL' is obtained from the old
volume WAL by shifting the volume WSW to the position
LSI/, through a horizontal distance hh '. Therefore the hori
zontal shift of the centre of gravity of the immersed volume
from its original position at B, or BR, is given by
v X hK
(using the principle discussed on p. 100). Therefore the
moment of statical stability at the angle 6 is
W ( V X y M  BG . sin (9 ) foottons
This is known as " Atwood's formula."
176 Theoretical Naval Architecture.
The righting arm or lever = ^~ BG . sin
If G is below B, as may happen in special cases
If we want to find the length of the righting arm or lever
at a given angle of heel 0, we must therefore know
(1) The position of the centre of buoyancy B in the up
right condition.
(2) The position of the centre of gravity G of the ship.
(3) The volume of displacement V.
(4) The value of the moment of transference of the wedges
parallel to the new waterline, viz. v X hH.
This last expression involves a considerable amount of cal
culation, as the form of a ship is an irregular one. The methods
adopted will be fully explained later, but for the present we
will suppose that it can be obtained when the form of the ship
is given.
Curve of Statical Stability. The lengths of GZ thus
obtained from Atwood's formula will vary as the angle of
heel increases, and usually GZ gradually increases until an
angle is reached when it obtains a maximum value. On
further inclination, an angle will be reached when GZ becomes
zero, and, further than this, GZ becomes negative when the
couple W X GZ is no longer a couple tending to right the
ship, but is an upsetting couple tending to incline the ship still
further. Take H.M.S. Captain l as an example. The lengths
of the lever GZ, as calculated for this ship, were as follows :
At 1 degrees, GZ = 4^ inches At 35 degrees, GZ = 7 inches
14 > " ~ "2 > 42 > ~ 5i
,, 21 ,, ,, io ,, 49 ,, ,, = 2 ,,
,, 28 ,, = 10 ,, ,, 54 = nil
Now set along a baseline a scale of degrees on a con
1 The Captain was a rigged turretship which foundered in the Bay of
Biscay. A discussion of her stability will be found in "Naval Science,"
vol. i.
Statical Stability, Curves of Stability, etc. 177
venient scale (say \ inch = i degree), and erect ordinates
at the above angles of the respective lengths given. If now
we pass a curve through the tops of these ordinates, we shall
obtain what is termed a "curve of statical stability" from
which we can obtain the length of GZ for any angle by drawing
the ordinate to the curve at that angle. The curve A, in Fig.
68, is the curve so constructed for the Captain. The angle
I
14. 21, 2a 35.
ANGLES OF
42. 49. 545.
INCLINATION.
FIG. 68.
at which GZ obtains its maximum value is termed the " angle of
maximum stability" and the angle at which the curve crosses
the baseline is termed the " angle of vanishing stability" and
the number of degrees at which this occurs is termed the
" range of stability" If a ship is forced over beyond the angle
of vanishing stability, she cannot right herself; GZ having a
negative value, the couple operating on the ship is an up
setting couple.
In striking contrast to the curve of stability of the Captain
is the curve as constructed for H.M.S. Monarch? The lengths
of the righting levers at different angles were calculated as
follows :
At 7 degrees, GZ = 4 inches
14
21
28
35
1 The Monarch was a rigged ship built about the same time as the
Captain^ but differing from the Captain in having greater freeboard. See
also the volume of "Naval Science " above referred to.
178 Theoretical Naval Architecture.
At 42 degrees, GZ = 22 inches
49 = 20
" == ~
The curve for this ship, using the above values for GZ, is
given by B, Fig. 68. The righting lever goes on lengthening
in the Monarch's case up to the large angle of 40, and then
shortens but slowly ; that of the Captain begins to shorten at
about 21 of inclination, and disappears altogether at 54^, an
angle at which the Monarch still possesses a large righting lever.
Referring to Atwood's formula for the lever of statical
stability at the angle 0, viz.
we see that the expression consists of two parts. The
first part is purely geometrical, depending solely upon the
form of the ship ; the second part, BG . sin 6, brings in the
influence of the position of the centre of gravity of the ship,
and this depends on the distribution of the weights forming
the structure and lading of the ship. We shall deal with these
two parts separately.
(1) Influence of form on curves of stability.
(2) Influence of position of centre of gravity on curves of
stability.
(i) We have here to take account of the form of the ship
above water, as well as the form of the ship below water. The
three elements of form we shall consider are draught, beam,
and freeboard. These are, of course, relative ; for con
venience we shall keep the draught constant, and see what
variation is caused by altering the beam and freeboard. For
the sake of simplicity, let us take floating bodies in the form
of boxes. 1 The position of the centre of gravity is taken as
constant. Take the standard form to be a box :
Draught ............... 21 feet.
Beam ............... 50^ ,,
Freeboard ...... ... ... ... 6J ,,
1 These illustrations are taken from a paper read at the Institution of
Naval Architects by Sir N. Barnaby in 1871.
Statical Stability , Ctirves of Stability, etc. 179
The curve of statical stability is shown in Fig. 69 by the
curve A. The deckedge becomes immersed at an inclination
of 14^, and from this angle the curve increases less rapidly
than before, and, having reached a maximum value, decreases,
the angle of vanishing stability being reached at about 38.
Now consider the effect of adding 4^ feet to the beam,
thus making the box
Draught 21 feet.
Beam 55
Freeboard ... ... ... ... ... 6J ,,
The curve is now given by B, Fig. 69, the angle of vanish
ing stability being increased to about 45. Although the
0. 10. 20. 30. 4O. 50. 60. 70.
ANGLE OF INCLINATION.
FIG. 69.
position of the centre of gravity has remained unaltered, the
increase of beam has caused an increase of GM, the meta
centric height, because the transverse metacentre has gone up.
We know that for small angles the lever of statical stability is
given by GM . sin 0, and consequently we should expect the
curve B to start as shown, steeper than the curve A, because
GM is greater. There is a very important connection between
the metacentric height and the slope of the curve of statical
stability at the start, to which we shall refer hereafter.
Now consider the effect of adding 4^ feet to the freeboard
of the original form, thus making the dimensions
Draught 21 feet.
Beam ... ... 50^ M
Freeboard II
180 Theoretical Naval Architecture.
The curve is now given by C, Fig. 69, which is in striking
contrast to both A and B. The angle of vanishing stability
is now 72. The curves A and C coincide up to the angle at
which the deckedge of A is immersed, viz. 14^, and then,
owing to the freeboard still being maintained, the curve C
leaves the curve A, and does not commence to decrease
until 40.
These curves are very instructive in showing the influence
of beam and freeboard on stability at large angles. We see
(a) An increase of beam increases the initial stability, and
therefore the slope of the curve near the origin, but does not
greatly influence the area enclosed by the curve or the range.
(b) An increase of freeboard has no effect on initial
stability (supposing the increase of freeboard does not affect
position of the centre of gravity), but has a most important
effect in lengthening out the curve and increasing its area.
The two bodies whose curves of statical stability are given by
A and C have the same GM, but the curves of statical stability
are very different.
(2) We now have to consider the effect on the curve of
statical stability of the position of the centre of gravity. If
the centre of gravity G is above the centre of buoyancy B, as is
usually the case, the righting lever is less than = by the
expression BG . sin 6. Thus the deduction becomes greater as
the angle of inclination increases, because sin increases as 6
increases, reaching a maximum value of sin 6 = i when =
90; the deduction also increases as the C.G. rises in the
ship. Thus, suppose, in the case C above, the centre of gravity
is raised 2 feet. Then the ordinate of the curve C at any
angle is diminished by 2 x sin 6. For 30, sin = , and
the deduction is there i foot. In this way we get the curve D,
in which the range of stability is reduced from 72 to 53 owing
to the 2feet rise of the centre of gravity.
It is usual to construct these curves as indicated, the
ordinates being righting levers, and not righting moments. The
righting moment at any angle can be at once obtained by
multiplying the lever by the constant displacement. The real
Statical Stability, Ctirves of Stability, etc. 181
curve of statical stability is of course a curve, the ordinates of
which represent righting moments. This should not be lost
sight of, as the following will show. In Fig. 70 are given the
15.
75.
curves of righting levers for a merchant vessel in two given
conditions, A for the light condition at a displacement of
1500 tons, and B for the load condition at a displacement of
3500 tons. Looking simply at these curves, it would be
thought that the ship in the light condition had the better
stability; but in Fig. 71, in which A represents the curve of
o.
75.
righting moments in the light condition, and curve B the curve
of righting moments in the load condition, we see that the
ship in the light condition has very much less stability than in
the load condition.
We see that the following are the important features of a
curve of statical stability :
(a) Inclination the tangent to the curve at the origin has to
the baseline ;
(b) The angle at which the maximum value occurs, and the
length of the righting lever at this angle ;
(c) The range of stability.
1 82 Statical Stability, Curves of Stability, etc.
The angle the tangent at the origin makes with the base
line can be found in a very simple manner as follows : At the
angle whose circular measure 1
is unity, viz. 57*3, erect a
perpendicular to the base,
and make its length equal to
the metacentric height GM,
for the condition at which
the curve has to be drawn,
using the same scale as for
the righting levers (see Fig.
72). Join the end of this
FIG. 72. .
line with the origin, and the
curve as it approaches the origin will tend to lie along this line.
The proof of this is given below. 2
Specimen Curves of Stability. In Fig. 73 are given
some specimen curves of stability for typical classes of ships.
A is the curve for a modern British battleship of about 3^
feet metacentric height. The range is about 63.
B is the curve for the American monitor Miantonomoh.
This ship had a low freeboard, and to provide sufficient stability
a very great metacentric height was provided. This is shown
by the steepness of the curve at the start.
C is the curve for a merchant steamer carrying a miscel
laneous cargo, with a metacentric height of about 2 feet. In
1 See p. 91.
2 For a small angle of inclination 0, we know that GZ = GM x 0,
6 being in circular measure ;
GZ GM
or ~0 = T
If now we express in degrees, say 6 = <f>, fhen
GZ GM
</> angle whose circular measure is I
G2
Z GM
If a is the angle OM makes with the base, then
GM GZ
tana = 5F3 = ^
and thus the line OM lies along the curve near the origin.
Statical Stability , Curves of Stability ', etc. 183
this ship there is a large righting lever even at 90. It must
be stated that, although this curve is typical for many ships, yet
10. 20. 30. 40. 50. 60. 70. 80. 90.
ANGLES OF INCLINATION.
FIG. 73.
the forms of the curves of stability for merchant steamers must
vary considerably, owing to the many different types of ships
and the variation in loading. Fig. 74 gives curves of stability
10
30 40 50 573 60 70
DEGREES OF INCLINATION
FIG. 74.
for several conditions of the T.S.S. Smolensk, 470' X 58'
X 37' (Mr. Rowell, I.N.A., 1905). They may be taken as
typical curves of a modern steamship of the highest class. A
is the curve for the load condition, in which the lower holds
are filled with 1200 tons of cargo, and the 'tween decks are
filled with 600 tons of cargo homogeneously stowed. All coal,
stores, and water are assumed on board. The GM is 1*5 feet,
and the range is 80. B assumes the cargo is all homo
geneous, the GM being reduced to i foot. The range is
rather over 70. C is for the same cargo as B, but all coal
1 84
Theoretical Naval Architecture.
is consumed except 200 tons in bottom of bunkers, and half
stores and fresh water only remain on board. The GM is
only o'6 foot, but the lighter draught has the effect of lengthen
ing out the curve to a range of yi^ . D is the condition
when the vessel is " light," having the ballast and reserve feed
tanks full; and the bunkers full of coal. The GM is 2 '8 feet,
and the range is over 90. It will be noticed that the
tangent at the origin has been drawn in each case at the angle
f" 1 TVT
a such that tan a = 5
57*3
The curves E and F in Fig. 74A have been prepared to
illustrate the effect of raising the centre of gravity of ship when
10
J
70
26 a ^ ___ j
FIG. 74A.
in condition C. If the centre of gravity is raised o'6 foot by
a different disposition of the cargo, the GM is zero, and the
curve of stability starts at a tangent to the base line. At all
angles the GZ is reduced from that in condition C by o  6
sin 0, so that we get the curve E, in which the range is 66.
If now we suppose the centre of gravity of ship lifted still
higher, viz. 05 foot, the vessel has a negative GM in the
upright condition, and is therefore in equilibrium, which is,
however, unstable. This is shown by the way the curve starts
at the origin at an angle of a = tan' 1 7. The ship will
heel until at 10 the centre of gravity and new CB again get
into the same vertical line and the ship is in equilibrium.
This time, however, the ship is in stable equilibrium, and has
a positive GM, so that the ship will loll over to 10, and there
be perfectly safe in calm weather, as is shown by the way the
curve of stability stretches out to a range of 62. At sea it
Statical Stability, Curves of Stability, etc. 185
would be advisable to fill up some of the ballast tanks, to
improve the stability of the vessel in either of the conditions
E and F.
Ships do occasionally get into the condition represented in
F, Fig. 74A. The following has reference to the S.S. Leo, which
capsized in 1895 (taken from Captain Owen's book "Aids to
Stability "). This ship left port with a cargo of barley and
wheat and 40 tons of coal on deck. Her freeboard was high,
no ballast being taken on board, and she had a list of 10 to
starboard or port, showing a negative GM. There was some
loose water in the bilges which the pump suctions could not
touch, as the ship had a list on one side or the other. The
ship listing to starboard, the engineers, to reduce the list, used
most of the coal from the starboard bunkers. This, however,
had the effect of raising the centre of gravity of the ship still
further, and increasing her instability when forced to the up
right. The wind and sea both now acted on the starboard
side, so that she returned to the upright and then lurched over
to port. The effect of the motion of the ship, and the force of
the wind sending the ship over to leeward, caused her to go
far beyond her natural position of equilibrium, say 10 to 15,
and this was helped by the loose water rushing across. Conse
quently such an angle was reached that the shifting boards
gave way and the grain got over to the inclined side, and the
ship went right over. The remedy in such a case would un
doubtedly have been to fill the waterballast tanks, so that the
ship had a positive GM in the upright condition.
A ship may start her voyage with a small positive GM, but,
owing to consumption of coal, etc., during the voyage, she may
get a list owing to a negative GM in the upright condition.
The most comfortable ship at sea is one with a small GM,
and if this is associated with such a position of the centre of
gravity and such a freeboard that the curve gives a good
maximum GZ and a good range, say like A in Fig. 74, we
have most satisfactory conditions of comfort and seaworthiness.
For smallcargo vessels it is generally recognized that the GM
should not be less than 0*8 foot, provided that a righting arm
of like amount is obtained at 30 to 40.
i86
Theoretical Naval Architecture.
For warships the conditions of stability are special. Here,
although the high freeboard is conducive to a good area of
the curve of stability and a large range, as seen in Fig. 69,
curve C, the conditions of design lead to a high position of
the centre of gravity, because of the disposition of guns and
armour. This discounts the effect of freeboard, as seen by
curve D in Fig. 69.
D, in Fig. 73, is the curve of stability for a sailingship
having a metacentric height of 3^ feet.
The curve of stability for a floating body of circular form fe
very readily obtainable, because
the section is such that the
upward force of the buoyancy
always acts through the centre
of the section, as shown in Fig.
75. The righting lever at any
angle 6 is GM . sin 0, where G
is the centre of gravity, and
M the centre of the section.
Taking the GM as two feet,
then the ordinates of the curve
of stability are o, 1*0, 173, 20,
I> 73> I<0 > at intervals of 30.
The maximum occurs at 90, and the range is 180. The
curve is shown in Fig. 76. A similar curve is obtained for a
submarine boat, the ordinate at angle being BG . sin 0, and
the range 180.
180.
Calculations for Curves of Stability. We now pro
ceed to investigate methods that are or have been adopted in
practice to determine for any given ship the curve of righting
levers. The use of the integrator is now very general for
Statical Stability, Curves of Stability , etc. 187
doing this, and it saves an enormous amount of work ; but, in
order to get a proper grasp of the subject, it is advisable to
understand the rrfethods that were in use previous to the intro
duction of the integrator.
In constructing and using curves of stability, certain assump
tions have to be made. These may be stated as follows :
1. The sides and deck are assumed to be watertight for the
range over which the curve is drawn.
2. The C.G. is taken in the same position in the ship, and
consequently we assume that no weights shift their position
throughout the inclination.
3. The trim is assumed to be unchanged, that is, the ship
is supposed to be constrained to move about a horizontal longi
tudinal axis fixed in direction only, and to adjust herself to the
required displacement without change of trim.
It is not possible in this work to deal with all the systems
of calculation that have been employed ; a selection only will
be given in this chapter. For further information the student
is referred to the Transactions of the Institution of Naval
Architects, and to the work by Sir E. J. Reed on the
11 Stability of Ships." The following are the methods that will
be discussed :
r. Blom's mechanical method.
2. Barnes' method.
3. Direct method (sometimes employed as a check on
other methods).
4. By Amsler's Integrator and Crosscurves of stability.
5. Tabular method (used at Messrs. John Brown).
6. Mr. Hok's method (given later).
i. Blom's Mechanical Method. Take a sheet of
drawingpaper, and prick off from the bodyplan the shape of
each equidistant section * (i.e. the ordinary sections for displace
ment), and cut these sections out up to the waterline at which
the curve of' stability is required, markirg on each section the
1 In settling the sections to be used for calculating stability by any of the
methods, regard must be had to the existence of a poop or forecastle the
ends of which are watertight, and the ends of these should as nearly as
possible be made stop points in the Simpson's rule.
1 88 Theoretical Naval Architecture.
middle line. Now secure all these sections together in their
proper relative positions by the smallest possible use of gum.
The weight of these represents the displacement of the ship.
Next cut out sections of the ship for the angle at which the
stability is required, taking care to cut them rather above
the real waterline, and gum together in a similar manner to
the first set. Then balance these sections against the first
set, and cut the sections down parallel to the inclined water
line until the weight equals that of the first set. When this
is the case, we can say that at the inclined waterline the
displacement is the same as at the original waterline in the up
right condition. This must, of course, be the case as the vessel
heels over. On reference to Fig. 67, it will be seen that what
we want to find is the line through the centre of buoyancy for
the inclined position, perpendicular to the inclined waterline,
so that if we can find B' for the inclined position, we can com
pletely determine the stability. This is done graphically by
finding the centre of gravity of the sections we have gummed
together, and the point thus found will give us the position of
the centre of buoyancy fon the inclined condition. This is done
by successively suspending the section ; and noting where the
plumblines cross, as explained on p. 5 1 . Having then the centre
of buoyancy, we can draw through it a line perpendicular to the
inclined waterline, and if we then spot off the position of the
centre of gravity, we can at once measure off the righting lever
GZ. A similar set of sections must be made for each angle about
10 apart, and thus the curve of stability can be constructed.
2. Barnes's Method of calculating Statical
Stability. In this method a series of tables are employed,
called Preliminary and Combination Tables, in which the work
is set out in tabulated form. Take the section in Fig. 77 to
represent the ship, WL being the upright waterline for the con
dition at which the curve of stability is required. Now, for a
small transverse angle of inclination it is true that the new water
plane for the same displacement will pass through the centre
line of the original waterplane WL, but as the angle of inclina
tion increases, a plane drawn through S will cut off a volume of
displacement sometimes greater and sometimes less than the
Statical Stability , Curves of Stability, etc. 189
original volume, and the actual waterline will take up some
such position as W'L', Fig. 77, supposing too great a volume to
be cut off by the plane through S. Now, we cannot say
straight off where the waterline W'L' will come. What we
have to do is this : Assume a waterline wl passing through
S ; find the volume of the assumed immersed wedge /SL, the
volume of the assumed emerged wedge wSW, and the area of
the assumed waterplane wl. Then the difference of the
volumes of the wedges divided by the area of the waterplane
will give the thickness of the layer between wl and the correct
waterplane, supposing the difference of the volumes is not too
great. If this is the case, the area of the new waterplane is
FIG. 77.
found, and a mean taken between it and the original. In this
way the thickness of the layer can be correctly found. If the
immersed wedge is in excess, the layer has to be deducted ; if
the emerged wedge is in excess, the layer has to be added.
To get the volumes of either of the wedges, we have to
proceed as follows : Take radial planes a convenient angular
interval apart, and perform for each plane the operation sym
bolized by ij> 2 . dx, t.e. the halfsquares of the ordinates
are put through Simpson's rule in a foreandaft direction for
each of the planes. Then put the results through Simpson's
rule, using the circular measure of the angular interval. The
Theoretical Naval Architecture.
result will be the volume of the wedge at the particular angle.
For proof of this see below. 1
' The results being obtained for the immersed and emerged
wedges, we can now determine the thickness of the layer. This
work is arranged as follows : The preliminary table, one table
for each angle, consists of two parts, one for the immersed
wedge, one for the emerged wedge. A specimen table is given
on p. 192 for 30. The lengths of the ordinates of each radial
plane are set down in the ordinary way, and operated on by
Simpson's multipliers, giving us a function of the area on the
immersed side of 550*3, and on the emerged side of 477*3.
We then put down the squares of the ordinates, and put them
through the Simpson's multipliers, giving us a result for the
immersed side of 17,878, and for the emerged side 14,251.
The remainder of the work on the preliminary table will be
described later.
We now proceed to the combination table for 30 (see
p. 193), there being one table for each angle. The functions
of squares of ordinates are put down opposite their respective
angles, both for the immersed wedge and the emerged wedge,
up to and including 30, and these are put through Simpson's
multipliers. In this case the immersed wedge is in excess, and
so we find the volume of the layer to be taken off to be 7839
cubic feet, obtaining this by using the proper multipliers. At
the bottom is placed the work necessary for finding the thickness
of the layer. We have the area of the whole plane 20,540
square feet, and this divided into the excess volume of the
immersed wedge, 7839 cubic feet, gives the thickness of the
layer to take off, viz. 0*382 foot, to get the true waterline.
We now have to find the moment of transference of the
1 The area of the section S/L is given by J ' y* . d6 y as on p. 15, and the
volume of the wedge is found by integrating these areas right fore and
aft, or
!/.*.*
which can be written
]//>*.'* .4
or f(lfy*.dx)a$
**' i fo* dx is found for each radial plane, and integrated with respect to
the angular interval.
Statical Stability, Curves of Stability, etc. 191
wedges, v X M' in Atwood's formula, and this is done by using
the assumed wedges and finding their moments about the line ST,
and then making at the end the correction rendered necessary
by the layer. To find these moments we proceed as follows :
In the preliminary table are placed the cubes of the ordinates of
the radial plane, and these are put through Simpson's rule ; the
addition for the emerged and immersed sides are added
together, giving us for the 30 radial plane 1,052,436. These
sums of functions of cubes are put in the combination table for
each radial plane up to and including 30, and they are put
through Simpson's rule, and then respectively multiplied by
the cosine of the angle made by each radial plane with the
extreme radial plane at 30. The sum of these products gives
us a function of the sum of the moments of the assumed im
mersed and emerged wedges about ST. The multiplier for the
particular case given is 0*3878, so that the uncorrected moment
of the wedges is 3,391, 336,* in footunits, i.e. cubic feet, multi
plied by feet.
1 The proof of the process is as follows : Take a section of the wedge
S/L, Fig. 78, and draw ST perpendicular to S/. Then what is required is
the moment of the section about ST, and this
integrated throughout the length. Take P and
P' on the curved boundary, very close together,
and join SP, SP' ; call the angle P'Sl, 6, and a.vc os ,
the angle P6P', d9. K* 2
Then the area PS P' = %y* . de SP y
The centre of gravity of SPP' is distant
from ST, \y . cos 0, and the moment of SPP'
about ST is
(iy > dQ ) x ^y . cos 0) FIG. 78.
or \y* . cos . dQ
We therefore have the moment of /SL about ST
\jy 3 . cos B . dQ
and therefore the moment of the wedge about ST is
J(jJ> 3 . cos 6 . d6)dx
or iffy 3 , cose .dx.de
i.e. find the value of ify 3 . cos 6 . dx for radial planes up to and including
the angle, and then integrate with respect to the angular interval. It will
be seen that the process described above corresponds with this formula.
PRELIMINARY TABLE FOR STABILITY.
WATER SECTION INCLINED AT 30.
IMMERSED WEDGE.
.
g
<_
**
"8
.
"S
!i
<5
(4
_C
is*
11
ii
J
Ii
*o
II
I
Ii
!
B
"g
H
II
*3
i
3d
us
1
o S
i
4' 5
j
2'2
20
J
IO
91
^
46
2
184
2
368
339
2
678
6,230
2
12,460
3
286
I
286
818
I
818
23,394
I
2 3,394
4
338
2
676
1142
2
2284
38,614
2
77,228
I
355
35'6
I
2
35'5
712
1260
1267
I
2
1260
2534
44,739
I
2
44,739
90,236
7
35'6
I
35'6
1267
I
1267
45! Il8
I
8
35'6
2
712
1267
2
2534
45, IlS
2
90,236
9
356
I
35'6
1267
I
1267
I
10
35 8
2
712
1267
2
2534
45!n8
2
90,236
ii
I
33'8
1142
I
1142
38,614
I
38,614
12
269
2
538
724
2
1448
19,465
2
38,930
13
H'3
*
72
205
*
102
2,924
i
1462
5503
17,878
597,8i7
EMERGED WEDGE.
I
47
*
23
22
J
II
104
i
52
2
13*8
2
27*6
I9O
2
380
2,628
2
5,256
3
214
I
214
458
I
458
9,800
I
9,800
4
27'3
2
546
745
2
1490
20,346
2
40,692
5
328
I
328
1076
I
1076
35,288
I
35,288
6
367
2
73*4
1347
2
2694
49,431
2
98,862
7
382
I
382
H59
I
1459
55,743
I
55,743
8
2
73'o
1332
2
2664
48,627
2
97,254
9
33'6
I
336
1129
I
1129
37,933
I
37,933
10
293
2
586
858
2
I7l6
25,154
2
ii
238
I
238
566
I
5 66
13,481
I
i3,'48i
12
169
2
33'8
286
2
573
4,827
2
9,654
13
8'4
i
42
7i
*
35
593
i
296
477*3
14,25!
Emerged 454,619
Immersed 597,817
1,052,436
1 The multipliers used here are half the ordinary Simpson's multipliers ;
the results are multiplied at the end by two to allow for this.
COMBINATION TABLE FOR STABILITY.
CALCULATION FOR GZ AT 30.
IMMERSED WEDGE.
o 5 xs
il.i
L
1*
Inclinations of
radial planes.
'S'SP
**!
2
^ 1 1 5
Sums of functi
of cubes of or
nates for bot
sides.
Multipliers.
Products of su
of functions of c
for both side
Cosines of incl
tions of radia
planes.
Functions of cu
for moments
wedges.
Function
ordinates
radial pla

!
Function
squares of
nates for vo
of wedg
___
15,340
i
15,340
974,388
I
974,388
08660
843,820
IO
15,760
4
63,040
990,153
4
3,960,612
09397
3,721,787
20
16,840
I 3
25,260
1,034,251
M
1,551,377
09848
1,527,796
25
30
550
17,701
17,878
2
i
35,402
8,939
1,066,771
1,052,436
2
2,133,542
526,218
0*9962
I 000
2,125,434
526,218
Immersed wedge 147,991
8,745,065
Emerged wedge 134,522
Multiplier f
03878
13,469 Uncorrected moment
3,391,336
Multiplier * 0582 Correction for layer
13,875
Volume of layer 7,839 cub. feet.
3,377,461
Volume of displacement
398,090
Longitudinal interval = 30 feet
BR
848
Circular measure 10 = 0*1745
BG sin 30 = 595
GZ = 253
* Multiplier = X 2 X (^ X 30) X (\ X 0*1745) = 0*582
f Multiplier = X 2 X ( X 30) X (\ X 0*1745) = 03878
BG = 1190; sin 30= 05 ; BG sin = 595 feet
EMERGED WEDGE.
AKEA AND POSITION OF C.G. OF RADIAL PLANE.
15,340
I
15,340
Functions Functions
of area. of moment.
IO
20
25
14^766
14,640
4
M
2
60,628
22,149
29,280
Immersed
Emerged
wedge 550
wedge 477
17,878
14,250
30
477
14.251
\
7,125
1027
3,628
134,522
Area =
1027 X 2 X
(i x 30)
= 20, 540 square feet
C.G. of radial plane on immersed side =
1027 5
1 77 feet
Thickness of layer =
7839 =Q .
382 foot
20540
NOTE. The work on the preliminary table may be much simplified by
using TchebycherFs sections : see Appendix A.
O
194 Theoretical Naval Architecture.
We now have to make the correction for the layer. We
already have the volume of the layer, and whether it has to be
added or subtracted, and we can readily find the position of
the centre of gravity of the radial plane. This is done at the
bottom of the combination table from information obtained on
the preliminary table. We assume that the centre of gravity
of the layer is the same distance from ST as the centre of
gravity of the radial plane, which may be taken as the case
unless the thickness of the layer is too great. If the layer is
thick, a new waterline is put in at thickness found, and the
area and C.G. of this waterline found. The mean between
the result of this and of the original plane can then be used.
The volume of the layer, 7839 cubic feet, is multiplied by the
distance of its centre of gravity from ST, viz. 177 feet, giving
a result of 13,875 in footunits, i.e. cubic feet multiplied by
feet. The correction for the layer is added to or subtracted
from the uncorrected moment in accordance with the following
rules :
If the immersed wedge is in excess, and the centre of gravity
of the layer is on the immersed side, the correction for the layer
has to be subtracted.
If the immersed wedge is in excess, and the centre of gravity
of the layer is on the emerged side, the correction for the layer
has to be added.
If the emerged wedge is in excess, and the centre of gravity
of the layer is on the emerged side, the correction for the layer
has to be subtracted.
If the emerged wedge is in excess, and the centre of gravity
of the layer is on the immersed side, the correction for the layer
has to be added.
These rules are readily proved, and are left as an exercise
for the student.
We, in this case, subtract the correction for the layer,
obtaining the true moment of transference of the wedges
as 3,377,461, or v X hh' in Atwood's formula. The volume
of displacement is 398,090 cubic feet; BG is 1190 feet;
sin 30 = 05. So we can fill in all the items in Atwood s
formula
Statical Stability, Curves of Stability, etc. 195
or GZ = 253 feet
In arranging the radial planes, it has been usual to arrange
that the deck edge comes at a stop point in Simpson's first rule,
because there is a sudden change of ordinate as the deck edge
is passed, and for the same reason additional intermediate
radial planes are introduced near the deck edge. In the case
we have been considering, the deck edge came at about 30.
The radial planes that were used were accordingly at
o, 10, 20, 25, 30, 35, 40, 50, 60, 70, 80, 90
These intermediate radial planes lead to rather complicated
Simpson's multipliers, and in order to simplify the calculations
it is thought to be sufficiently accurate to space the radial
planes rather closer, say every 9. For such a series of radial
planes the multipliers for 9, 18, 27, 36, 54, 72, 81, and 90
are readily obtained by one of Simpson's rules. For 45 the
multipliers can be 0*4, i, i, i, I, 0*4, with the multiplier ~ x
OT57. For 63 they can be i, f, f, f, 071, i, i, f, with the
multiplier 0*157. These can be readily proved ; 0*157 is the
circular measure of 9.
Barnes's method of calculating stability has been very largely
employed. It was introduced by Mr. F. K. Barnes at the Insti
tution of Naval Architects in 1861, and in 1871 a paper was
read at the Institution by Sir W. H. White and the late Mr.
John, giving an account of the extensions of the system, with
specimen calculations. For further information the student is
referred to these papers, and also to the work on " Stability,"
by Sir E. J. Reed. At the present time it is not used to any
large extent, owing to the introduction of the integrator,
which gives the results by a mechanical process in much less
time. It will be seen that in using this method to find the
stability at a given angle, we have to use all the angles up to
and including that angle at which the stability is required.
Thus a mistake made in the table at any of the smaller angles
is repeated right through, and affects the accuracy of the
1 96
Theoretical Naval Architecture.
calculation at the larger angles. In order to obtain an indepen
dent check at any required angle, we can proceed as follows :
3. Triangular or Direct Method of calculating
Stability. Take the bodyplan, and draw on the trial plane
through the centre of the upright waterline at the required
angle. This may or may not cut off the required displace
ment. We then, by the ordinary rules of mensuration, dis
cussed in Chapter I., find the area of all such portions as S/L,
Fig. 77, for all the sections, 1 and also the position of the centre
of gravity, g, for each section, thus obtaining the distance S//.
This is done for both the immersed and emerged wedges. The
work can then be arranged in tabular form thus
Number of
section.
Areas.
Simpson's
multipliers.
Products for
volume.
Levers as
&*,
Products for
moment about
ST.
I
2
A,
A,
I
4
A,
4 A,
*!
x*
A,*,
4A 2 * 2
etc.
etc.
etc.
etc.
etc.
etc.
S, M,
The volume of the wedge = B! X \ common interval
The moment of wedge about ST = Mj X \ common interval
This being done for both wedges, and calculating the area
of the radial plane, we can find the volume of the layer and the
uncorrected moment of the wedges. The correction for the
layer is added or subtracted from this, exactly as in Barnes's
method, and the remainder of the work follows exactly the
methods described above for Barnes's method.
The check spot at 90 is very readily obtained, because the
volume and C.G. of the portion above the L.W.L. are readily
determined, and we already know the volume and C.B. of
portion below the L.W.L. from the displacement sheet. If V
is the volume of displacement, and Vj is onehalf the volume
V
above the L.W.L., then  + V x is the volume when at 90 the
ship is immersed to the middleline plane. The C.G. of this
1 The sections are made into simple figures, as triangles and trapeziums,
in order to obtain the area and position of C.G. of each.
Statical Stability, Ciirves of Stability, etc. 197
volume is readily obtainable. The difference between this and
V
V will give the volume of layer, V\ = V 2 , say, where the
layer has to be added. The C.G. of this layer is readily
determined, as it will very nearly be that of the middleline
plane of the ship, so that the C.G. of the volume V is found
at once, and this gives the GZ at 90.
There is the disadvantage about the methods we have
hitherto described, that we only obtain a curve of stability for
one particular displacement, but it is often necessary to know
the stability of a ship at very different displacements to the
ordinary load displacement, as, for example, in the light con
dition, or the launching condition. The methods we are now
about to investigate enable us to determine at once the curve
of stability at any given displacement and any assumed position
of the centre of gravity.
4. Amsler's Integrator. Crosscurves of Stability.
The Integrator is an extension of the instrument we have
described on p. 81, known as the planimeter. A diagram of
one form of the integrator is given in Fig. 79. A bar, BB, has
a groove in it, and the instrument has two wheels which run in
this groove. W is a balance weight to make the instrument
run smoothly. There are also three small wheels that run on
the paper, and a pointer as in the planimeter. By passing the
pointer round an area, we can find
(1) A number which is proportional to the area^ i.e. a
function of the area.
(2) A function of the moment of the area about the axis
the bar is set to.
(3) A function of the moment of inertia of the area about
the same axis.
The bar is set parallel to the axis about which moments
are required, by means of distance pieces.
(1) is given by the reading indicated by the wheel marked A.
(2) is given by the reading indicated by the wheel marked M.
(3) is given by the reading indicated by the wheel marked I.
The finding of the moment of inertia is not required in our
present calculation.
I 9 8
Theoretical Naval Architecture.
Now let M'LMW represent the bodyplan > of a vessel
inclined to an angle of 30; then, as the instrument is set, the
FIG. 79.
axis of moments is the line through S perpendicular to the
inclined waterline, and is what we have termed ST. What
we want to find is a line through the centre of buoyancy in the
inclined position perpendicular to the inclined waterline. By
passing the pointer of the instrument round a section, as
W'L'M, we can determine its area, and also its moment about
the axis ST by using the multipliers ; and doing this for all the
sections in the body, we can determine the displacement and
also the moment of the displacement about ST. 2 Dividing the
moment by the displacement, we obtain at once the distance
of the centre of buoyancy in the inclined condition from the
axis ST. It is convenient in practice to arrange the work in
a similar manner to that described for the pi ani meter, p. 83,
and the following specimen calculation for an angle of 30 will
illustrate the method employed. 3 Every instrument has multi
pliers for converting the readings of the wheel A into areas,
and those of the wheel M into moments. The multipliers
must also take account of the scale used.
1 The bodyplan is drawn for both sides of the ship the forebody in
black, say, and the afterbody in red.
2 This is the simplest method, and it is the best for beginners to employ ;
but certain modifications suggest themselves after experience with the
instrument. See Example 23 in this chapter.
* See Appendix A for calculation, using " TchebychefFs rule " with the
integrator.
Statical Stability, Ciirves of Stability, etc. 199
AREAS.
MOMENTS.
SECTIONS.
j
I
ss
d
)
bo
J
"d
2
1
i
I
G ^
.5*3
1
1
1
1
3
H
Q
M
Q
W g
Initial readings
I and 17
3,210
6 4
i
6 4
3900
3910
10
i
10
2,4,6,8, 10, 12, 14, and 16
3 5> 7, 9, ii, !3> and J S
8,859
14,345
5649
5486
4
2
22,596
10,972
3124
2381
+ 7*6
+ 743
4
2
3144
1486
33,632 4620
Multiplier for displacement = 0*02
Multiplier for moment = 0*2133
Displacement = 33,632 X 0*02 = 672*6 tons
Moment = 4620 x 0*2133 = 985 foottons
GZ = 6^6= r 4 6feet
In this case the length of the ship was divided into sixteen
equal parts, and accordingly Simpson's first rule can be
employed. The common interval was 8*75 feet. The multi
plier for the instrument was y^ for the areas, and $ for the
moments, and, the drawing being on the scale of J inch =
i foot, the readings for areas had to be multiplied by (4)2 = 16,
and for moments by (4) 3 = 64. The multiplier for displace
ment in tons is therefore
; 16 x ( X 8*75) x^ = 002
and for the moment in foottons is
li$o X 64 X ( X 875) X ^ = 02133
We therefore have, assuming that the centre of gravity is at S
GZ = 2 ^ = 1*46 feet
6726
Now, this 672*6 tons is not the displacement up to the
original waterline WL, and we now have to consider a new
conception, viz. crosscurves of stability. These are the converse
200
Theoretical Naval Architecture.
of the ordinary curves of stability we have been considering.
In these we have the righting levers at a constant displacement
and varying angles. In a crosscurve we have the righting
levers for a constant angle, but varying displacement. Thus
in Fig. 79, draw a waterline W"L" parallel to W'L', and for
CROSS CURVES or STABILITY.
C.G.IMLW.L.
3000.
TONS
4000.
DISPLACEMENT.
SOOO
FIG. 80.
the volume represented by W"ML" find the displacement and
position of the centre of buoyancy in exactly the same way as
we have found it for the volume WML'. The distance which
this centre of buoyancy is from the axis gives us the value of
GZ at this displacement, supposing the centre of gravity is at
S. The same process is gone through for two or more water
lines, and we shall have values of GZ at varying displacements
at a constant angle. These can be set off as ordinates of a
curve, the abscissae being the displacements in tons. Such a
curve is termed the " crosscurve of stability " at 30, and for
any intermediate displacement we can find the value of GZ at
30 by drawing the ordinate to the curve at this displacement.
A similar process is gone through for each angle, the same
position for the centre of gravity being assumed all through,
Statical Stability, Curves of Stability, etc. 201
and a series of crosscurves obtained. Such a set of cross
curves is shown in Fig. 80 for displacements between 3000
and 5000 tons at angles of 15, 30, 45, 60, 75, and 90. At
any intermediate displacement, say at 4600 tons, we can draw
the ordinate and measure off the values of GZ, and so obtain
the ordi nates necessary to construct the ordinary curve of
stability at that displacement and assumed position of the
centre of gravity. The relation between the crosscurves and
the ordinary curves of stability is clearly shown in Fig. 81.
1,500
IS. 30.
DEGREES
45. 60. 75.
OF INCLINATION.
90.
FIG. 81.
We have four curves of stability for a vessel at displacements
of 1500, 2000, 2500, and 3000 tons. These are placed as
shown in perspective. Now, through the tops of the ordinates
at any given angle we can draw a curve, and this will be the
crosscurve of stability at that angle.
It will have been noticed that throughout our calculation
we have assumed that the centre of gravity is always at the
point S, and the position of this point should be clearly stated
on the crosscurves. It is evident that the centre of gravity
cannot always remain in this position, which has only been
assumed for convenience. The correction necessary can
readily be made as follows : If G, the centre of gravity, is
below the assumed position S, then GZ = SZ f SG . sin 0, and
2O2 Theoretical Naval Architecture.
if G is above S, then GZ = SZ  SG . sin for any angle 6.
Thus the ordinates are measured from the crosscurves at the
required displacement, and then, SG being known, SG sin 1 5,
SG sin 30, etc., can be found, and the correct values of GZ
determined for every angle.
If an integrator is not available, crosscurves can be calcu
lated by using a modification of Barnes's tables already discussed.
Three poles, O, O', O", are taken, and Barnes's tables are worked
out for each set of radial planes through these poles ; but no
correction is made for the layer, as for crosscurves we set off
the lever for whatever the displacement comes to (see Example
22). For each angle there are thus three spots obtained, and
by the method described below tangents are drawn at each of
these places to the curves. At 90 the result found for each
pole should be the same; at 72 and 81, say, the spots come
very close together and do not give reliable curves. A separate
calculation is therefore made for 90 over a wide range of dis
placement, which can readily be done, and then curves for
constant displacement are run in fair, so that auxiliary spots on
the crosscurves at 72 and 81 are obtained.
5. Tabular Method of calculating Stability. The
following method of obtaining a crosscurve is a very con
venient one to employ ; the whole of the work being arranged
in tabular form and Tchebycheff's rule being used, the fore
and aft integration is easily performed by addition. Take the
complete body of a ship, inclined as shown in Fig. Si A, the
axis AA being perpendicular to the new waterline, and prefer
ably always taken through the same spot in the middle line,
say, the intersection of the M.L. and the L.W.L. It is the
practice at Messrs. John Brown and Co., Clydebank, to use
this method as an independent check against the results of the
Integrator, and the tabular form reproduced in Table IV. 1 is
given by the kind permission of Mr. W. J. Luke. Tcheby
cheff s 10 ordinate rule is used for the fore and aft integration.
In the case given, the waterlines are placed 3' 6" apart, the
lowest one being tangential to the lowest point of the bilge.
For each waterline the moment of the area of waterplane
1 See at the end of the book.
Statical Stability, Ciirves of Stability, etc. 203
about A A is J[jy. dxj(y'f. dx\, and the area of the waterplane
is \Jy . dx\jy'. dx\. Thus, for No. 4 W.L. the sum of ordinates
port and starboard side is 2965, and "this is a function of the
area of the plane. As regards moment about A A, the starboard
side sum of squares is in excess of the port side by 980,
which is therefore a function of the moment of the plane on
the starboard side. These functions are converted into area
and moment, giving 15,860 square feet and 26,216 (square feet
X feet). These are obtained in the first part of the table for
all the waterlines considered necessary. In the second portion,
the areas and moments of waterlines are integrated vertically
to 2, 4, 6, 8, 10 waterlines successively as shown. The sum
of functions of areas are turned into tons displacement ; thus, at
2 W.L. : displacement in tons = 19,875 X 5 X 3*5 X ^ = 662
tons. The sum of functions of moments are turned into
moment; thus, at 2 W.L. moment in foot tons = 165,735 x \
X 3*5 X 3*5 = 5524. The division 5524^662 = 835 feet,
which gives the distance of the C.B. of the displacement to
No. 2 W.L. on the starboard side of A A, and thus one spot
on the crosscurve at 30 is obtained, viz. 8*35 feet, at a dis
placement of 662 tons. If the moment on the port side is in
excess for any W.L., that is then made negative. Thus, for
Nos. 6, 7, 8 in the table the port side is in excess. Similarly,
when integrating the waterlines forward and aft, if the section
crosses the axis AA, the ordinate is given a negative value. This,
of course, will occur more frequently as the angle gets larger.
Tangent to a Crosscurve. If, as is usual, the cross
curves are drawn to represent righting levers on base of tons
displacement, the tangent at any point on a crosscurve has an
7 (~* >7
inclination 6 to the base line given by tan 6 = '  If now
M is the righting moment in foottons, GZ = ^ so that
: ,TT T l TTT ) ^y ^^y ^y 2 ~ \V\. ^W ~" W
2O4 Theoretical Naval Architecture.
Now, </M is the increment of moment due to increment of
layer <AV, so that  is the distance of the C.G. of radial
plane from ST = S^, say. Therefore tan 6 = ^(S^  GZ).
Sg is the same distance from ST as the centre of gravity of the
radial plane we are dealing with, cutting off the displacement
at which we are drawing the tangent.
If r lt r z are ordinates of radial plane on immersed and
emerged sides respectively, then with our usual notation
S(r, + rjdx
on the immersed side. These can readily be picked up on
the Barnes's tables. If S^ > GZ, and g is to the right of S,
then tan 6 is positive ; if S^ is to the left of S, tan is negative.
If SG is < GZ, tan 6 is negative. These tangents must be set
off, having in view the scales that are used for righting levers
and for displacement.
Dynamical Stability. The amount of work done by
a force acting through a given distance is measured by the
product of the force and the distance through which it acts.
Thus, a horse exerting a pull of 30,000 Ibs. for a mile does
30,000 X 1760 X 3 = 158,400,000 footlbs. of work
Similarly, if a weight is lifted, the work done is the product of
the weight and the distance it is lifted. In the case of a
ship being inclined, work has to be done on the ship by some
external forces, and it is not always possible to measure the
work done by reference to these forces, but we can do so by
reference to the ship herself. When the ship is at rest, we
have seen that the vertical forces that act upon the ship are
(1) The weight of the ship acting vertically downwards
through the centre of gravity ;
(2) The buoyancy acting vertically upwards through the
centre of buoyancy ;
these two forces being equal in magnitude. When the ship is
Statical Stability, Curves of Stability, etc. 205
inclined, they act throughout the whole of the inclination.
The centre of gravity is raised, and the centre of buoyancy is
lowered. The weight of the ship has been made to move
upwards the distance the centre of gravity has been raised, and
the force of the buoyancy has been made to move downwards
the distance the centre of buoyancy has been lowered. The
work done on the ship is equal to the weight multiplied by the
rise of the centre of gravity added to the force of the buoyancy
multiplied by the depression of the centre of buoyancy ; or
Work done on the ship = weight of the ship multiplied by the
vertical separation of the centre of gravity and the centre
of buoyancy.
This calculated for any given angle of inclination is termed
" the dynamical stability " at that angle, and is the work that
has to be expended on the ship in heeling her over to the
given angle.
Moseley's Formula for the Dynamical Stability
at any Given Angle of Inclination. Let Fig. 67 repre
sent a vessel heeled over by some external force to the
angle 0; g, g' being the centres of gravity of the emerged
and immersed wedges ; gk, g'h' being drawn perpendicular to
the new waterline W'L f . The other points in the figure have
their usual meaning, BR and GZ being drawn perpendicular
to the vertical through B'.
The vertical distance between the centres of gravity and
buoyancy when inclined at the angle 6 is B'Z.
The original vertical distance when the vessel is upright
is BG.
Therefore the vertical separation is
B'Z  BG
and according to the definition above
Dynamical stability = W(B'Z  BG)
where W = the weight of the ship in tons. *
Now, B'Z = B'R + RZ = B'R + BG . cos
Now, using v for the volumes of either the immersed wedge
206 Theoretical Naval Architecture.
or the emerged wedge, and V for the volume of displace
ment of the ship, and using the principle given on p. 100,
we have
Substituting in the above value for B'Z, we have _
th : t sr al
which is known as Mo s eley* s formula.
It will be seen that this formula is very similar to Atwood's
formula, and it is possible to calculate it out for varying
angles by using the tables in Barnes' method of calculating
stability. It is possible, however, to find the dynamical
stability of a ship at any angle much more readily if the
curve of statical stability has been constructed, and the
method adopted, if the dynamical stability is required, is as
follows :
The dynamical stability of a ship at any given angle
is equal to the area of the curve of statical
stability up to that angle (the ordinates of this curve
being the actual righting moments).
Referring to Fig. 67, showing a ship heeled over to a cer
tain angle 0, imagine the vessel still further heeled through a
very small additional angle, which we may call dB. The centre
of buoyancy will move to B" (the student should here draw his
own figure to follow the argument). B'B" will be parallel to
the waterline W'L', and consequently the centre of buoyancy
will not change level during the small inclination. Drawing a
vertical B"Z' through B", we draw GZ', the new righting arm,
Statical Stability, Ciirves of Stability, etc. 207
perpendicular to it. Now, the angle ZGZ' = dB, and the ver
tical separation of Z and Z' = GZ x dB. Therefore the work
done in inclining the ship from the angle B to the angle B j dd
is
W x (GZ . dB)
Take now the curve of statical stability for this vessel. At
the angle B the ordinate is GZ. Take a consecutive ordinate
at the angle B + dQ. Then the area of such a strip = GZ x dB \
but this multiplied by the displacement is the same as the
above expression for the work done in inclining the vessel
through the angle dB t and this, being true for any small angle
dB, is true for all the small angles up to the angle B. But the
addition of the work done for each successive increment of
inclination up to a given angle is the dynamical stability at
that angle, and the sum of the areas of such strips of the curve
of statical stability as we have dealt with above is the area of
that curve up to the angle B. Therefore we have the dynamical
stability of a ship at any given angle of heel is equal to the
area of the ordinary curve of statical stability up to that angle,
multiplied by the displacement.
To illustrate this principle, take the case of a floating body
whose section is in the form of a circle, and which floats with
its centre in the surface of the water. The transverse meta
centre of this body must be at the centre of the circular section.
Let the centre of gravity of the vessel be at G, and the centre
of buoyancy be at B. Then for an inclination through 90
G will rise till it is in the surface of the water, but the centre of
buoyancy will always remain at the same level, so that the
dynamical stability at 90 = W x GM.
Now take the curve of statical stability for such a vessel.
The ordinate of this curve at any angle = VV x GM . sin 0,
and consequently the ordinates at angles 15 apart will be
W . GM . sin o, W . GM . sin 15, and so on ; or 0258
W.GM, 05 W.GM, 0707 W.GM, 0866 W . GM, 0965
W . GM, and W.GM. If this curve is set out, and its area
calculated, it will be found that its area is W x GM, which is
the same as the dynamical stability up to 90, as found above.
2o8 Theoretical Naval Architecture.
It should be noticed that the angular interval should not be
taken as degrees, but should be measured in circular measure
(see p. 90). The circular measure of 15 is 02618.
The dynamical stability at any angle depends, therefore,
on the area of the curve of statical stability up to that angle ;
and thus we see that the area of the curve of stability is of
importance as well as the angle at which the ship becomes un
stable, because it is the dynamical stability that tells us the
work that has to be expended to force the ship over. For full
information on this subject the student is referred to the
" Manual of Naval Architecture," by Sir W. H. White, and Sir
E. J. Reed's work on the " Stability of Ships."
Mr. Hok's Method of obtaining a Curve of Sta
bility. In this method the ordinary planimeter is used, and
as the use of curves through various spots obtained is a feature
of the method, the work is readily checked as one proceeds.
The method first obtains the curve of dynamical stability for
a given displacement, and then from this curve the curve of
statical stability is deduced. The former curve is the integral
of the latter, and so the latter is the differential of the former.
That is, if H is the dynamical stability at angle 6, then
.GZ.</0 and GZ = ~ ~
Take the body prepared with the sections on both sides in
the ordinary way inclined, as shown in Fig. SIA. Then, by
means of the planimeter, we can determine the displacement
up to the various dotted waterlines, and so construct a curve
of displacement. A line parallel to the base line and distance
away equal to the displacement V, is drawn as shown, and this
gives the draught at which we shall cut off the displacement
required, and for which we desire the righting lever. The
area of owl divided by wl gives the distance of the C.B. below
W.L., so that wb = owl^wl. The area owl is readily
obtained by the planimeter. Now, G being the assumed
position of the centre of gravity, and B' the new C.B., B being
the C.B. when upright, the dynamical stability = W(B'ZBG),
and B'Z BG can be readily found by measurement. By
Statical Stability, Curves of Stability, etc. 209
repeating this for a number of angles, a curve of dyna
mical stability can be drawn, observing that the W portion
can be left out, being constant all through. If h be such
that H = W ./#, and ^ 10 , AJOJ ^o> etc., be the values at 10,
20 } 3> etc > an d GZ 10 , GZao, etc., be the values of the righting
arm at 10, 20, etc., then at 10 we have
o = A (o'i745)(5 X o + 8 x
(0*1745 . > the c.m. of 10)
n = i(o'i745)(o + 4 X GZ^ +
and
By these two equations GZ 10 and GZ 20 can be obtained. Also
Ao = f (oi745)(o + 3 GZ 10 + 3 GZso + GZ 30 ), and GZ 10 and GZ^
being known GZ 30 is found. Thus values of GZ can be deter
mined and the ordinary curve of stability drawn in for the
given displacement and assumed C.G. (For all practical
purposes ^ 10 = JGZ 10 X 0*1745, seeing that the ordinary curve
of stability is straight near the origin.) This can be done for
other assumed displacements, and so a set of crosscurves
drawn in for constant angles on a base of displacement, as
already explained. All these, of course, assume a constant
position for the C.G., and if at any displacement another
position of the C.G. has to be allowed for, GG' sin is added
or subtracted, according as the new G' is below or above the
oldG.
2IO Theoretical Naval Architecture.
Stability of Selfrighting Lifeboats. The stability of
these boats offers several points of interest. The properties of
such boats are
(1) Very large watertight reserve of buoyancy, which
renders the boat practically unsinkable.
(2) The water shipped is automatically cleared.
(3) The loss of stability due to shipping water is not
sufficient to cause instability.
(4) The boat is unstable when upside down.
(1) For this the ends of the boat have great sheer, and the
ends are filled in with aircases or tanks. These cases are also
placed under the deck and at the sides between the deck and
seats (see Fig. SIB). The cases are in such numbers that even
if some are broached there would still be a sufficient buoyant
volume left.
The large buoyant volume at the ends gives great lifting
power to the boat when encountering a sea.
(2) The deck of the boat is somewhat higher than the
waterline, and in this deck, passing through to the bottom, are
eight tubes with automatic freeing valves t which allow water to
pass down, but not up. These are adjusted so as to drop
down with a small pressure of water above. The rise and
fall of the boat will soon cause any water on the deck to
be discharged.
The deck falls towards midships, and has a " round down,"
so that water will flow to the valves.
(3) Shipping water on the deck has two effects on the
stability, viz.
(a) Will raise the C.G. due to added weight
(b) Will make the virtual C.G. higher than the actual C.G
with the added water, because of the free surface
(Rise of G = where / is the moment of inertia of
free surface, and V is volume displacement with
added water).
In order that these two effects may not render the boat
unstable
*i.) An iron keel is fitted to pull C.G. down j
Statical Stability, Curves of Stability, etc.
21 I
212
Theoretical Naval Architecture.
(ii.) The ends remain intact, so that the lost moment of
inertia i shall not be too great.
The GM in this extreme condition should be positive, or
otherwise when the boat, after capsizing, came back to the up
right she would again capsize, owing to the water on the deck.
This GM will, however, only be small, and as the water
level falls the i will reduce considerably owing to the presence
of the side tanks, and the water becoming less in quantity also
helps matters.
FIG. 8ic.
(4) This instability when upside down is the property
known as selfrighting. In this condition, let B', M' be C.B.
and metacentre respectively, and G the centre of gravity.
Then G must be nearer the keel than M'.
The buoyancy being provided largely by the ends, B' will
be a good way from keel, G is also near keel owing to iron
keel ; therefore B'G is large. Again, B'M' is not large, because
the moment of inertia is provided largely by the ends ; therefore
we get M' below G, and the boat is unstable.
It is also necessary that the boat should have a curve of
stability as ABC, Fig. 8ic, the only positions of equilibrium
being at upright and 180. If the curve of stability were like
ADFC, there would be a position of stable equilibrium at F,
which would not be desirable. If the curve were as AEC,
then the equilibrium at 180 would be stable.
These boats are tested, when fully equipped, with mast up
and sail set, by immersing to the gunwale by weights to repre
sent men. The boat is then turned upside down by a chain
attached to a crane. The chain is then slipped, when the boat
should return to the upright position.
Statical Stability, Curves of Stability, etc. 213
In Fig. SIB the boat is of the type with a drop keel, and
provision is made for water ballast. There are eight automatic
freeing valves.
Stability of Sailing Vessels Power to carry sail.
For comparative purposes the sail area is taken as the "plain"
or " working " sail, and this is assumed all braced fore and
aft. This sail for a ship would include " jib," " fore and main
courses," " driver," three "topsails," and three "topgallant
sails." The centre of effort vs. assumed as the C.G. of the area
of these sails. The centre of lateral resistance of the water is
taken as the C.G. of the middle line area. The couple heeling
the ship is caused by the resultant of the wind pressure and
the fluid pressure on the opposite side of the ship. If A is
area of sails in square feet, p is pressure in Ibs. per square
foot, h the vertical distance between the C.E. and the C.L.R.,
and 6 the angle of heel, then the moment of the couple heeling
A y A \^s 7j
the ship is  , and this equals the moment of stability
W x GM X sin 0. Taking for comparative purposes a pressure
of i Ib. per square foot (equivalent to a wind of about fourteen
knots).
i W x GM
^InT = Ax h X22 *
This is termed the power to carry sail, and is a measure of the
stiffness of a ship. The greater this is, the less a given vessel
will incline under a given wind. Sailing merchant vessels
have a value 12 to 20 ; the sloops in the Navy, 12 to 15 ; smaller
values are usual in sailing yachts, as the crew have an influence
on the heel by going over to the leeward side.
If a curve of wind moment be constructed on base of
angle and plotted on the curve of stability (ordinates repre
senting righting moments), where the two cross will represent
where the stability equals the heeling moment, and this is the
angle of steady heel in Fig. 8 ID this is 10. The area of sail
projected on to the vertical plane is A x cos 6, and the lever
of the moment is h x cos 0, so that the heeling moment is
p . A . h . cos 2 6, and from this a curve can be drawn as AD,
2I 4
Theoretical Naval Architecture.
Fig 8 ID. If now the ship is supposed upright and exposed
to a sudden squall, the work done by the wind will be the area
10
G2O 30 40
FIG. SID.
50
60
under the wind moment curve. The work done to any angle
by the stability will be the area under the curve of stability.
At the angle of steady heel, the former is in excess by the
amount OAB, and the ship will heel over until the area OAB
is equal to the area BEF. If the wind remains constant, the
ship will eventually settle to the angle of steady heel in this
case 10. The area BEDF above the wind curve is thus the
reserve the vessel has against further wind pressure, and this
area is termed the " reserve of dynamical stability"
If a ship is struck by a squall at the moment of complet
ing a roll to windward, say 10, the wind moment and the
stability of the ship both act together in taking the vessel to
the upright, the work being represented by the area OAEF,
G
FIG. 81
Fig. 8 IE. It is only after passing the upright that the stability
acts against the wind, and the ship must heel to an angle 29,
such that the area FEABO is equal to the area BCG. It is
thus seen that a sailing vessel with a low curve of stability, like
Statical Stability, Curves of Stability, etc. 215
the Captain (Fig. 68), may have insufficient reserve of dynamical
stability, and under the above circumstances, might be blown
right over. Mr. Wall, I.N.A.,
1914, introduced this principle
into ordinary ship calculations.
Heel produced by Gun
Fire 1 (Fig. 8 IF). This problem
has to use the principle of
momentum. If w and W be
the weights of projectiles (and
powder) and the ship respec
tively, v the velocity of the pro Vs  ] ^
jectile, then on firing, the C.G. FIG. SIP.
of ship will have a backward
velocity of V, and if I is the impulsive reaction of the water
at rather less than half draught, we have the equation
w W
I = .z/  .V.
g g
 . v being the momentum of the shot, and .V that of the ship.
A o
The angular momentum of the ship is . & . , and this
o
has to be equated to the moment of momentum causing the
rotation or
from which
V is practically negligible, so that
w W dd
ATJ
.v. AH = .k* . r.
g g dt
I 72
= *x/
A/ m.
from which k z can be found.
g
/jf)\%
The initial kinetic energy of the ship is J . W . k* .( j j , and
if 6 is the angle of heel this is equated to the dynamical
1 See Chap. IX. on The Rolling of Ships for the definitions of T and /*.
216
Theoretical Naval Architecture.
stability at 6 supposing resistances are neglected, i.e.
J . W . m . 2 ,
regarding the curve of stability a straight line.
v . AH\ 2
z/.AH
w v . AH.TT
! W' m.g.T
circular measure
As an example, take a case in which 8 guns are fired on the
broadside 25 feet above water, H being 13 feet below water,
w = 1 100 Ibs., velocity of discharge 3000 fs., G.M. 5 feet,
displacement 18,000 tons, T = 8 seconds.
.. . , noo X 3000 X 38 X 8 X 180
6 in degrees = 224O x l8>ooo x 5 x 32  2 x 8
= 3j degrees nearly.
Example. Determine the heel caused by firing simultaneously 4 guns
at a muzzle velocity of 1600 feet per second, the weight of projectile and
powder being 2375 Ibs., the height of guns above centre of lateral
resistance being 30 ft., the time of a single oscillation 6 seconds, and the
metacentric height 4^ feet. Displacement of ship 10,000 tons.
A us. 4^ degrees.
Angle of Heel of a Vessel when Turning. On put
ting a vessel's rudder over the pressure on the rudder which
acts below the centre of lateral
resistance tends
vessel inwards.
FIG. SIG.
to heel the
This inward
heeling is very noticeable in the
case of destroyers, in which the
rudder area is relatively large.
In ordinary ships this is only
of short duration, and when the
the vessel gets on the circle, an outward heel is caused by the
centrifugal action, which acts above the centre of lateral
resistance (C.L.R.). The moment caused by the product of
the centrifugal force and the distance of the C.G. from the
C.L.R. is the moment causing the heel, and this is equated to
the moment of stability at angle 6. This is determined.
The centrifugal force Q caused by a weight W tons moving
Statical Stability, Curves of Stability , etc. 217
W z/ 2
in a circle radius R feet at speed v feet per second is
8 R
W # 2
and the moment of the couple causing heel is =? d. This
is equated to the moment of stability at angle 6, viz. W. GM
sin 6, or
sn =
= ' 88 (irds) (v in knots) 
The features therefore which lead to a large heel when turning
are (i) high speed, (2) small turning circle, and (3) small
metacentric height.
Example. A vessel whose tactical diameter is 463 yds. at a point on
her circular turn has a speed of 15 knots, the draught being 27 ft. and the
metacentric height 35 ft. Approximate to the angle of heel.
In this case V = 15, R = 695 ft. d  13 ft. about. GM = 3^ ft.
/. sin 6 = 0088 . ^ . 12 = 0106, and 6 = 6.
Metacentric Height when inclined about an Axis
inclined at an Angle a with the longitudinal Middle
Line Plane. We first have to
find the moment of inertia of
the waterplane about an axis
inclined to the principal axes
OX and OY, viz. OZ in Fig. SIR.
O is the C.G. of waterplane.
Drawing as in figure.
I = 2SAxDQ 2 whereSA
is an element of
area
= 2SA(j cos a x sin a) 2
FIG. 8iH.
cos 2 a + x 2 sin 2 a 2xy sin a cos a)
The last term vanishes on summation since the axes are
through the C.G. of plane.
.. I = cos 2 a/f . dK + sin 2 a/* 2 . dh
I T . cos 2 a + It sin 2 a
.'. BM tt = BM T . cos 2 a + BM L . sin 2 a
also BG = BG . (cos 2 a + sin 2 a) since cos 2 a f sin 2 a = i
/. GM a = GM, . cos 2 a + GM L . sin 2 a.
218 Theoretical Naval Architecture.
Example. A boxshaped vessel is 80' long, 20' wide and floats at a
draught of water of 10'. Find the value of the distance between the
centre of buoyancy and the metacentre for inclinations about an axis
coincident with a diagonal of the rectangular waterplane. (Honours B. of
E. 1911).
In this case
a = tan 1 \
sin 2 a = T ' 7 , cos 2 a = j$
BM T = 3'33, BMj = 5333
/. BM a = (333 X tf) + (5333 X T ' 7 ) = 627 feet.
EXAMPLES TO CHAPTER V.
1. A twomasted cruiser of 5000 tons displacement has its centre of
gravity at two feet above the waterline. It is decided to add a military
top to each mast. Assuming the weight of each military top with its guns,
men, and readyammunition supply to be 12 tons, with its centre of gravity
70 feet above the waterline, what will be the effect of this change on
(1) The metacentric height of the vessel ?
(2) The maximum range of stability, assuming the present maximum
range is 90, and the tangent to the curve at this point inclined
at 45 to the baseline ?
(Scale used, \ inch = i, \ inch = ^ foot GZ.)
Ans. (i) Reduce 0325 foot, assuming metacentric curve horizontal ;
(2) reduce range to about 86^, assuming no change in cross
curves from 5000 to 5024 tons.
2. The curve of statical stability of a vessel has the following values of
GZ at angular intervals of 15 : o, 055, i'O3, 099, 0*66, 024, and 0*20
feet. Determine the loss in the range of stability if the C.G. of the ship
were raised 6 inches.
Ans. 16.
3. Obtain, by direct application of Atwood's formula, the moment of
stability in foottons at angles of 30, 60, and 90, in the case of a prismatic
vessel 140 feet long and 40 feet square in section, when floating with sides
vertical at a draught of 20 feet, the metacentric height being 2 feet.
4. A body of square section of 20 feet side and 100 feet long floats with
one face horizontal in salt water at a draught of 10 feet, the metacentric
height being 4 inches. Find the dynamical stability at 45.
Ans. 171 foottons.
5. Indicate how far a vessel having high bulwarks is benefited by them
as regards her stability. What precautions should be taken in their con
struction to prevent them becoming a source of danger rather than of safety ?
6. Show from Atwood's formula that a ship is in stable, unstable, or
neutral equilibrium according as the centre of gravity is below, above, or
coincident with the transverse metacentre respectively.
7. A vessel in a given condition displaces 4600 tons, and has the C.G.
in the 19feet waterline. The ordinates of the crosscurves at this
displacement, with the C.G. assumed in the igfeet waterline, measure as
follows: 0*63, 138, 215, 2'o6, 137, 056 feet at angles of 15, 30,
45, 60, 75, and 90 respectively. The metacentric height is 24 feet.
Draw out the curve of stability, and state (i) the angle of maximum
stability, (2) the angle of vanishing stability, and (3) find the dynamical
stability at 45 and 90.
Ans. (i) 5of : (2) iooi; (3) 3694, 9650 foottons.
Statical Stability, Curves of Stability, etc. 219
8. A vessel has a metacentric height of 34 feet, and the curve of stability
has ordinates at 15, 30, 37^, 45, and 60 of 09, 1*92, 2*02, 165, and
0*075 feet respectively. Draw out this curve, and state the angle of
maximum stability and the angle at which the stability vanishes.
Ans. 351, 59^
9. A vessel's curve of stability has the following ordinates at angles of
15. 3o 45, 60, and 75, viz. 051, 097, 090, 053, and 0*08 feet
respectively. Estimate the influence on the range of stability caused by
lifting the centre of gravity of the ship o'2 foot.
Ans. Reduce nearly 6.
10. A square box of 18 feet side floats at a constant draught of 6 feet,
the centre of gravity being in the waterline. Obtain, by direct drawing or
otherwise, the value of GZ up to 90 at say 6 angles. Draw in the curve
of statical stability, and check it by finding its area and comparing that
with the dynamical stability of the box at 90.
(Dynamical stability at 90 = 3 X weight of box.)
11. A vessel fully loaded with timber, some on the upper deck, starts
from the St. Lawrence River with a list. She has two crossbunkers extend
ing to the upper deck. She reaches a British port safely, with cargo undis
turbed, but is now upright. State your opinion as to the cause of this.
12. Show by reference to the curves of stability of boxshaped vessels
on p. 174 that at the angle at which the deck edge enters the water the
tangent to the curve makes the maximum angle with the baseline.
13. The curve of stability of a vessel at angles of 15, 30, 45, 60,
75, and 90 shows the following values of the righting arm, viz. 0*22, 071,
105, 102, 0*85, and 0*56 feet respectively, the metacentric height being
8 inches and the displacement being 4500 tons. Discuss in detail the
condition and behaviour of the ship if 200 tons were removed from a hold
17 feet below the centre of gravity.
(Assume that the crosscurves from 4300 to 4500 Ions are all parallel to
the baseline.']
14. A vessel of 1250 tons displacement has its centre of gravity loj feet
above keel. The stability curve (on scale \ inch = i, \ inch ^ foot)
ends as a straight line at 20 slope to the base line, the range being 80.
Find the alteration in metacentric height and range of stability due to
taking in 30 tons reserve feed water 3 feet above the keel.
Ans. Increase 0*176 foot in GM.
,, 5 in range.
15. Show that for a wallsided vessel inclined to an angle
GZ = sin 6 (GM + JBM . tan 2 0)
where GM and BM refer to the upright condition.
1 6. Show, by using the above formula, that if a wallsided vessel has a
negative metacentric height she will loll over to an angle (J> such that
17. Apply the answer to question 16 to show that the log shown
floating in Fig. 55 with a negative G.M. of , will take up the position
corner downwards if left free.
1 8. A boxshaped vessel is 100 feet long, 30 feet broad, and 16 feet
deep. In the load condition the freeboard is 4 feet and the metacentric
height is 6 feet. In the light condition the freeboard is 10 feet and the
220
Theoretical Naval Architecture.
metacentric height is still 6 feet. Compare fully the stability of the vessel
in these two conditions. (This should be treated in the light of remarks
on p. 181.)
19. A vessel 72 feet long floats at 6 feet draught and has 4^ feet free
board, with sides above water vertical. Determine the GZ at 90, the C.B.
when upright being 3$ feet above keel, the C.G. I foot below waterline,
the halfordinates of waterplane being O'8, 3*3, 5*4, 65, 6'8, 6*3, 5*1, 2'8,
o'6, and the displacement 100 tons. Ans. + o 67 foot.
20. A prismatic vessel is 32 feet broad, 13 feet draught, 9 feet freeboard,
the bilges being circular arcs of 6 feet radius. GM is 2 feet.
(1) Obtain the first part of curve of stability by formula in question
15 above.
(2) Obtain values of GZ at 45 and 72 by using Barnes's method, using
9 angular intervals.
(3) Obtain GZ at 90 by direct method.
Thus draw in the complete curve of stability.
21. In the above vessel, instead of the sides above water being vertical,
they fall in from I foot above the waterline to the deck, where the breadth
is 24 feet. Obtain the complete curve of stability in a similar manner to
the preceding question.
(The curves in these two questions are given in the author's textbook
on " Warships," chap, xix.)
22. In obtaining crosscurves by calculation as described above, if v,
v' are volumes of emerged and immersed wedges, between the upright
waterplane and the radial plane through O, and^, A, g y h\ are as in our
ordinary notation, show that the righting lever for displacement V + v' v
is given by
vx
 V.OB sing
V+z/'  v
so that, not needing to correct for layer to get the displacement V, we get
for example of Barnes's method in this chapter a lever of 2^52 feet at a
displacement of 10,158 tons and angle 30.
23. In using the integrator for stability calculation, the "figure 8"
method is frequently employed. This consists in using the displacement
MWL with its C.B. B as a basis, and running round sections of wedges in
direction SL'L, WW'S. By this means the integrator adds up the
FIG. 8xi.
moments of wedges and subtracts the volumes. If w, itf are displacements
of in and out wedges, the displacement to W'L' is W + (area reading X
Statical Stability, Curves of Stability, etc. 221
proper factor). The GZ at angle 6 and displacement W 4 w  w' is
given by
(moment reading X proper factor)  (W . ES . sin 6)
{W + (area reading X proper factor)}
24. A boxshaped vessel 420 feet long, 72 feet broad, and 24 feet
constant draught has a compartment amidships 60 feet long, with a W.T.
middle line bulkhead extending the whole depth. Determine the angle of
heel caused by the vessel being bilged on one side abreast this bulkhead,
the C.G. of the vessel being 23 feet above the keel.
To what height should the transverse bulkheads at the ends of the
bilged compartment be carried, so as to confine the water to this part of
the vessel ?
This is done in two steps : (i) sinkage, (2) heel. See Fig. 81 1.
volume lost buoyancy 60 X 24 X 36 _ e
Bodi* smkage = area intact w f P . = 39O x 72 = ' 8 5 feet '
I of intact W.P. about middlej = A . 36o . 72 , + j . fo . 36' = 12, 130, 560
C.F. from middle line as also\ _ . f .
the C.B. / 
Io Ih'ough CLF: ab Ut ""I = I2 ' I3 ' 56 ~ (39 ' 72 ' lT22)
= 12, 102, 560.
RiSC u f bdl ? g ne  half j = 092 feet.
the bodily sinkage /
,,,. 12 . 102 . 560 , ,. .
New B'M' =  =167 feet.
420 x 72 X 24
Vertical distance between Gj _ , fi
and M' before heeling / ~ I2 4 07230
= 6*62 feet.
The vessel must heel until G and M' are in the same vertical, so that,
6 being the angle of heel,
tan 6 = ?7 = O'l6 .*. 6 = 9 nearly.
The height of bulkhead = 240 + 1*85 + (3712 X tan 0)
= 318 feet nearly.
25. Example of a similar nature for a box 400 feet by 75 feet by 26 feet,
midship compartment with a M.L. bulkhead 50 feet long. C.G. of ship
25 feet from keel.
Ans. 6 = 12^, height of bulkhead, 36^23 feet.
26. Example of a similar nature for a box 350 feet by 60 feet by 20 feet,
midship compartment 35 feet with a M.L. bulkhead. G.M. = 8 feet.
Ans. = 6, height of bulkhead 24*19 feet.
27. A prismatic vessel 100 ft. long has a transverse section formed of a
rectangle, height 10 ft. breadth 20 ft. resting on the top of a semicircle of
radius 10 ft. The centre of gravity is 3 ft. above the keel and the draught
of water is 10 ft. Find the volume of the correcting layer and the
righting moment when the vessel is inclined 45, the displacement being
constant.
(B. of E. 1911.)
This is an excellent example to show the application of
222
Theoretical Naval Architecture.
Barnes' method, and the solution is accordingly given here
with. (Fig. 8ij).
Taking angular intervals of 15 we have the following for
FIG. 8ij.
the first part of the combination tables. The preliminary
tables are not necessary, seeing that the section of vessel is
constant.
IMMERSED WEDGE.
EMERGED WEDGE.
Incli
nation
of
radial
Ordi
nates.
Squares.
S.M.
Pro
ducts.
plane.
O
10
100
I
100
IS
I0'3
1 06
3
318
30
"5
132
3
396
45
1414
200
i
2OO
Ordi
nates.
Squares.
S.M.
Products.
10
100
I
IOO
IO
100
3
300
10
100
3
300
10
IOO
1
IOO
1014
800
Layer = (1014 800) X ^ X f X 0*2617
X 100 cubic ft.
2i4X3Xo'26i7Xioo
=   ^ =io5ocub.ft.
Area of radial plane = 100 X 2414.
Thickness of layer = 0435 ft
The table for finding the uncorrected moment of the
wedges is as follows : the cubes of the ordinates, on immersed
Statical Stability , Curves of Stability, etc. 223
and emerged sides of each radial plane being added together
before putting through the necessary multipliers to satisfy
.cvsO.dx. dO (see p. 191).
Angle of
radial
plane.
Sums of cubes
of ordinates
S.M.
Products.
Cos0
(from radial
plane).
Products.
2,000
r
2,OOO
0707
1,414
15
2,093
3
6,279
0866
5,440
3
2,521
3
7,563
0966
7,300
45
3,828
i
3,828
I'O
3,828
17,982
Uncorrected moment = 17,982 X i X f X 0*2617 x 100
= 58,750 cubic ft. x feet.
The C.G. of the layer is on the immersed side, 207 ft., so that
the layer correction is 1050 x 2*07 = 2170 cubic ft. x feet.
The layer has to be subtracted, seeing that immersed wedge
is in excess. The moment is accordingly deducted. The
corrected moment is therefore 58,750 2170 = 56,580, which
is the v X hK in At wood's formula.
V = 100 x ~ 7 . 100 .\ = 15,700 cubic ft.
B below W.L. = .radius = 425 ft.
GZ =
.'. BG is 275 ft.
yh + BG sin 0. (G below B.)
= 36 f 194 = 554 feet.
Righting moment =
o j
x 5'54 = 24,800 foot tons.
CHAPTER VI.
CALCULATION OF WEIGHTS STRENGTH OF BUTT
CONNECTIONS DAVITS, PILLARS, DERRICKS, AND
SHAFT BRACKETS.
Calculations of Weights. We have discussed in
Chapter I. the ordinary rules of mensuration employed in find
ing the areas we deal with in ship calculations. For any
given uniform plate we can at once determine the weight
if the weight per square foot is given. For iron and steel
plates of varying thicknesses, the weight per square foot is
given on p. 38. For iron and steel angles and y bars of
varying sizes and thicknesses tables are calculated, giving the
weight per lineal foot. Such a table is given on p. 225 for
steel angles, etc., the thicknesses being in ygths of an inch. It
is the Admiralty practice to specify angles, bars, etc., not in thick
ness, but in weight per lineal foot. Thus an angle bar 3" x 3"
is specified to weigh 7 Ibs. per lineal foot, and a Z bar 6" X
3 i" x 3" is specified to weigh 15 Ibs. per lineal foot. When the
bars are specified in this way, reference to tables is unnecessary.
The same practice is employed with regard to plates, the thick
ness being specified as so many pounds to the square foot.
If we have given the size of an angle bar and its thick
ness, we can determine its weight per foot as follows : Assume
the bar has square corners, and is square at the root, then, if
a and b. are the breadth of the flanges in inches, and / is the
thickness in inches, the length of material / inches thick in the
section is (a f b /) inches, or feet ; and if the bar
is of iron, the weight per lineal foot is
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226 Theoretical Naval Architecture.
If the bar is of steel, the weight per lineal foot is
X 4 ' 8 X / lbs *
Thus a 3" X 3" X f" steel angle bar would weigh 717 lbs.,
and a steel angle bar 3" x 3" of 7 lbs. per foot would be
slightly less than f inch thick.
It is frequently necessary to calculate the weight of a
portion of a ship's structure, having given the particulars of
its construction ; thus, for instance, a bulkhead, a deck,
or the outer bottom plating. In any case, the first step must
be to find the area of plating and the lengths of angle bars.
The weight of the net area of the plating will not give us the
total weight of the plating, because we have to allow for butt
straps, laps, rivetheads, and in certain cases liners. The method
employed to find the allowance in any given case is to take a
sample plate and find what percentage the additions come to
that affect this plate, and to use this percentage as an addition
to the net weight found for the whole. To illustrate this, take
the following example :
A deck surface of 10,335 square feet is to be covered with T 5 s inch
steel plating, worked flush, jointed with singleriveted edges and butts
Find the weight of the deck, allowing 3 per cent, for rivetheads.
steel plates are 1275 lbs. per square foot, so that the ne
fgi
ight
Q.335 x. 275 _ 88 tons
2240
Now, assume an average size for the plates, say 1 6' X 4'. finch rivet
will probably be used, and the width of the edge strip and butt strap wil
be about 5 inches. The length round half the edge of the plate is 20 feet
and the area of the strap and lap belopging to this plate is
20 X & = 8 33 square feet
The percentage of the area of the plate is therefore
X ioo = 13 per cent
04
Adding also 3 per cent, for rivetheads, the total weight is 68 '4 tons.
It is usual to add 3 per cent, to allow for the weight of rivet
heads. For lapped edges and butt straps, both double riveted,
Calculation of Weights, etc. 227
the percentage 1 comes to about 10 per cent, for laps, 5^ percent,
for butt straps, and 3 per cent, for liners as ordinarily fitted to
the raised strakes of plating. No definite rule can be laid
down, because the percentage must vary according to the
particular scantlings and method of working the plating, etc.,
specified.
The length of stiffeners or beams required for a given area
can be very approximately determined by dividing the area in
square feet by the spacing of the stiffeners or beams in feet.
For wood decks, 3 per cent, can be added for fastenings.
Example. The beams of a deck are 3 feet apart, and weigh 22 Ibs.
per foot run ; the deck plating weighs 10 Ibs. per square foot, and this is
covered by teak planking 3 inches thick. Calculate the weight of a part
54 feet long by 10 feet wide of this structure, including fastenings.
(S. and A. Exam. 1897.)
Net area of deck = 54 x 10 = 540
Add for butts and laps 7 per cent. = 3 7 '8
.5778
(Assume singleriveted butt straps and singleriveted laps.)
Weight of plating = 577*8 X 10
= 5778 Ibs.
Running feet of beams = ^ = 180
Weight of beams = 180 x 22
= 3960 lbs.
Total weight of plating and beams = 9, 738 Ibs
Add 3 per cent, for rivetheads = 292 ,,
10,030
Weight of teak 3 = 540 X % = 7155 Ibs.
Add 3 per cent, for fastenings = 215 ,,
Weight of wood deck 7370 ,,
Summary.
Plating and beams 10,030 Ibs.
Wood deck 7,370
Total ... 17,400 = 7'8 tons.
1 A number of percentages worked out for various thicknesses, etc,
will be found in Mr. Mackrow's " Pocket Book."
2 No allowance made for beam arms, which should be done if a whole
deck is calculated.
a Teak taken as 53 Ibs. per cubic foot.
228
Theoretical Naval Architecture.
Use of Curves. For determining the weight of some of
the portions of a ship, the use of curves i found of very great
assistance. Take, for instance, the transverse framing of a ship.
For a certain length this framing will be of the same character,
as, for example, in a battleship, within the double bottom,
where the framing is fitted intercostally between the longi
tudinals. We take a convenient number of sections, say the
sections on the sheer drawing, and calculate the weight of the
complete frame at each section. Then along a base of length
set up ordinates at the sections, of lengths to represent the
calculated weights of the frames at the sections. Through the
spots thus obtained draw a curve, which should be a fair line.
The positions of the frames being placed on, the weight of each
frame can be obtained by a simple measurement, and so the
total weight of the framing determined. The curve AA in
Fig. 82 gives a curve as constructed in this way for the transverse
framing below armour in the double bottom of a battleship.
Before and abaft the double bottom, where the character of the
framing is different, curves are constructed in a similar manner.
Weight of Outer Bottom Plating. The first step
necessary is to determine the area we have to deal with. We
FIG. 8a.
can construct a curve of girths, as BB, Fig. 82 ; but the area given
by this curve will not give us the area of the plating, becaus
although the surface is developed in a transverse directk
Calculation of Weights, etc. 229
there is no development in a longitudinal direction. (Strictly
speaking, the bottom surface of a ship is an undevelopable
surface.) The extra area due to the slope of the level lines is
allowed for as follows : In plate I., between stations 3 and 4,
a line fg is drawn representing the mean slope of all the level
lines. Then the ordinate of the curve of girths midway
between 3 and 4 stations is increased in the ratio fg : h. This
done all along the curve will give us a new modified curve of
girths, as B'B', Fig. 82, and the area given by this curve will give
a close approximation to the area of the outer bottom of the
ship. This is, of course, a net area without allowing for butt
straps or laps. Having a modified curve of girths for the
whole length, we can separate it into portions over which the
character of the plating is the same. Thus, in a vessel built
under Lloyd's rules, the plating is of certain thickness for one
half the length amidships, and the thickness is reduced before
and abaft. Also, in a battleship, the thickness of plating is the
same for the length of the double bottom, and is reduced
forward and aft. The curves A A and BB, Fig. 82, are con
structed as described above for a length of 244 feet.
Weight of Hull. By the use of these various methods,
it is possible to go right through a ship and calculate the
weight of each portion of the structure. These calculable
portions for a battleship are
(1) Skinplating and plating behind armour.
(2) Inner bottom plating.
(3) Framing within double bottom, below armour, behind
armour, and above armour. Outside double bottom, below
and above the protective deck.
(4) Steel and wood decks, platforms, beams.
(5) Bulkheads.
(6) Topsides.
There are, however, a large number of items that cannot be
directly calculated, and their weights must be estimated by
comparison with the weights of existing ships. Such items
are stem and stern posts, shaft brackets, engine and boiler
bearers, rudder, pumping and ventilation arrangements, pillars,
paint, cement, fittings, etc.
230 Theoretical Naval Architecture.
It is, however, a very laborious calculation to determine
the weight of the hull of a large ship by these means; and
more often the weight is estimated by comparison with the
ascertained weight of existing ships. The following is one
method of obtaining the weight of steel which would be used in
the construction of a vessel : The size of the vessel is denoted
by the product of the length, breadth, and depth, and for known
ships the weight of steel is found to be a certain proportion of
this number, the proportion varying with the type of ship.
The coefficients thus obtained are tabulated, and for a new ship
the weight of steel can be estimated by using a coefficient
which has been obtained for a similar type of ship. The weight
of wood and outfit can be estimated in a similar manner.
Another method is described by Mr. J. Johnson, M.I.N.A.,
in the Transactions of the Institution of Naval Architects for
1897, in which the sizes of vessels are represented by Lloyd's
old longitttdinal number* modified as follows : In threedecked
vessels, the girths and depths are measured to the upper deck
1 Lloyd's numbers (now superseded by the New Rules)
1. The scantlings and spacing of the frames, reversed frames, and floor
plates, and the thickness of bulkheads are regulated by nun.bers, which are
produced as follows :
2. For one and two decked vessels, the number is the sum of the
measurements in feet arising from the addition of the halfmoulded breadth
of the vessel at the middle of the length, the depth from the upper part of
the keel to the top of the upperdeck beams, with the normal roundup,
and the girth of the half midship frame section of the vessel, measured from
the centre line at the top of the keel to the upperdeck stringer plate.
3. For threedeck steam vessels, the number is produced by the
deduction of 7 feet from the sum of the measurements taken to the top of
the upper deck beams.
4. For spardecked vessels and awningdecked steam vessels, the
number is the sum of the measurements in feet taken to the top of the main
deck beams, as described for vessels having one or two decks.
5. The scantlings of the keel, stem, stern frame, keelson, and stringer
plates, the thickness of the outside plating and deck ; also the scantlings
of the angle bars on beam stringer plates, and keelson and stringer angles
in hold, are governed by the longitudinal number obtained by multiplying
that which regulates the size of the frames, etc., by the length of the vessel.
The measurements for regulating the above scantling numbers are taken
as follows :
I . The length is measured from the after part of the stem to the fore part
of the sternpost on the range of the upperdeck beams in one, two, and
three decked and spardecked vessels, but on the range of main deck beams
in awningdecked vessels.
In vessels where the stem forms a cutwater, the length is measured from
Calculation of Weights, etc. 231
without deducting 7 feet. In spar and awningdeck vessels,
the girths are measured to the spar or awning decks respec
tively. In one, two, and welldecked vessels, the girths and
depths are taken in the usual way. Curves are drawn for each
type of vessel, ordinates being the weight of iron or steel in
tons for vessels built to the highest class at Lloyd's or Veritas,
and abscissae being Lloyd's longitudinal number modified as
above. These curves being constructed for ships whose weights
are known, it is a simple matter to determine the weight for a new
ship of given dimensions. For further information the student
is referred to the paper in volume 39 of the Transactions.
To calculate the Position of the Centre of Gravity
of a Ship. We have already seen in Chapter III. how to find
the C.G. of a completed ship by means of the inclining experi
ment, and data obtained in this way are found very valuable in
estimating the position of the C.G. of a ship that is being
designed. It is evident that the C.G. of a ship when com
pleted should be in such a position as to obtain the metacentric
height considered necessary, and also to cause the ship to float
correctly at her designed trim. Suppose, in a given ship, the C.G.
of the naked hull has been obtained from the inclining experi
ment (that is, the weights on board at the time of the experi
ment that do not form part of the hull are set down and their
positions determined, and then the weight and position of the
C.G. of the hull determined by the rules we have dealt with in
Chapter III.). The position of the C.G. of hull thus determined
is placed on the midship section, and the ratio of the distance
of the C.G. above the top of keel to the total depth from the
top of keel to the top of the uppermost deck amidships will
the place where the upperdeck beam line would intersect the after edge of
stem if it were produced in the same direction as the part below the
cutwater.
2. The breadth in all cases is the greatest moulded breadth of the vessel.
3. The depth in one and two decked vessels is taken from the upper
part of the keel to the top of the upperdeck beam at the middle of the
length, assuming a normal roundup of beam of a quarter of an inch to a
foot of breadth. In spardecked vessels and awningdecked vessels, the
depth is taken from the upper part of the keel to the top of the maindeck
beam at the middle of the length, with the above normal roundup of
beam.
232 Theoretical Naval Architecture.
give us a ratio that can be used in future ships of similar type
for determining the position of the C.G. of the hull. Thus, in
a certain ship the C.G. of hull was 203 feet above keel, the
total depth being 34*4 feet. The above ratio in this case is
therefore 059, and for a new ship of similar type, of depth 395
feet, the C.G. of hull would be estimated at 395 Xo'59, or 233
feet above the keel. For the foreandaft position, a similar ratio
may be obtained between the distance of the C.G. abaft the
middle of length and the length between perpendiculars. In
formation of this character tabulated for known ships is found
of great value in rapidly estimating the position of the C.G.
in a new design.
For a vessel of novel type, it is, however, necessary to cal
culate the position of the C.G., and this is done by combining
together all the separate portions that go to form the hull.
Each item is dealt with separately, and its C.G. estimated as
closely as it is possible, both vertically and in a foreandaft
direction. These are put down in tabular form, and the total
weight and position of the C.G. determined.
In estimating the position of the C.G. of the bottom plating,
we proceed as follows : First determine the position of the
C.G. of the several curves forming the halfgirth at the various
stations. This is not generally at the halfgirth up, but is some
where inside or outside the line of the curve. Fig. 83 represents
the section AB at a certain station. The curve is divided into
four equal parts by dividers, and the C.G. of each of these parts
is estimated as shown. The centres of the first two portions
are joined, and the centres of the two top portions are joined
as shown. The centres of these lastdrawn lines, g^ g^ are
joined, and the centre of the line g^g^ viz. G, is the C.G. of
the line forming the curve AB, and GP is the distance from the
L.W.L. This done for each of the sections will enable us to
put a curve, CC in Fig. 82, of distances of C.G. of the half
girths from the L.W.L. 1 We then proceed to find the C.G. of
the bottom plating as indicated in the following table. The
area is obtained by putting the halfgirths (modified as already
1 This assumes the plating of constant thickness. Plates which are
thicker, as at keel, bilge, and sheer, can be allowed for afterwards.
Calculation of Weights, etc.
233
explained) through Simpson's rule. These products are then
multiplied in the ordinary way to find the foreandaft position
of the C.G. of the plating, and also by the distances of the C.G.
Fio. 83.*
of the sections below the L.W.L., which distances are measured
off from the curve CC and are placed in column 6. The
remainder of the work does not need any further explanation.
CALCULATION FOR AREA AND POSITION OF C.G. OF BOTTOM PLATING
FOR A LENGTH OF 244 FEET.
Modified
half girths.
Simpson's
multipliers.
Products.
Multiples
for leverage.
Products.
C.G. from
L.W.L.
Products.
41'5
SI''
I
4
415
2044
2
I
830
2044
181
216
751
4,415
53'0
2
io6'o
O
2874
22'2
2,354
49 6
37'5
4
1984
37'5
2
1984
75'o
213
I90
4,226
712
5878
2734
140
12,458
1 The C.G. of wood sheathing, if fitted, can be obtained from this
figure by setting off normally to the curve from G the halfthickness of the
sheathing.
234
Theoretical Naval Architecture.
Common interval = 61 feet
Area both sides = 587*8 x \ X 61 x 2
= 23,904 square feet
C.G. abaft middle of length of plating =  = x 61
5575
= 1*45 feet
C.G. below L.W.L. = Q >45 * = 212 feet
5878
CALCULATION FOR THE POSITION OF THE C.G. OF A VESSEL.
FROM L.W.L.
FROM MIDDLE OF LENGTH.
IM.BI

Below.
Above.
Before.
Abaft.
1TBMB.
s
_.
.
J
.
i

t>

B
5
w
s
1
3
%
*
*
Equipment
Water
25
40
100
I2'0
300
Provisions
3
45
135
250
75o
Officers' stores
15
2'0
30
I250
i,875
Officers,men,and effects
30
60
i so
5501650
Cables
30
40
120
8502550
Anchors
10
15*0
150
900
900
Masts, yards, etc.
25
450
1125
7'0
175
Boats
IO
21'0
210
20'0
200
Warrant officers' stores
20
15
30
650
1300
Armament
*75
4
700
5*o
875
Machinery
450
40
1800
14,850
Engineers' stores
So
'S
25
700
3,500
Coals
300
0'2
60
3*0
900
Protective deck
2IO
i 5
315
150
3,150
Hull
1250
__
"^
i5
1875
~
^~~
"'5
14,375
Total 2630 2300 4555 8050 39,300
tons 2300 8,050
2630)2255
086 ft.
above L.W.L
C.G. above L.W.L. = O'86
Trans, met. ,, =297
Trans, met. above C.G. = 2*97 0*86
= 2 ' 1 1 feet.
2630)31,250
1 1 88 ft.
abaft mid.
length.
Calculation of Weights, etc. 2 35
Calculation for C.G. of a Completed Vessel. By the
use of the foregoing methods we can arrive at an estimate of
the weight of hull, and also of the position of its C.G. relative
to a horizontal plane, as the L.W.P., and to a vertical athwart
ship plane, as the midship section. To complete the ship for
service, there have to be added the equipment, machinery, etc.,
and the weights of these are estimated, as also the positions of
their centres of gravity. The whole is then combined in a
table, and the position of the C.G. of the ship in the completed
condition determined.
The preceding is such a table as would be prepared for a
small protected cruiser. It should be stated that the table is
not intended to represent any special ship, but only the type of
calculation.
The total weight is 2630 tons, and the C.G. is 0*86 foot
above the L.W.L. and 11*88 feet abaft the middle of length.
The sheer drawing enables us to determine the position of the
transverse metacentre, and the estimated GM. is found to be
2'ii feet. The centre of buoyancy calculated from the sheer
drawing should also be, if the ship is to trim correctly, at a
distance of 1 1 '88 feet abaft the middle of length.
Strength of Butt Fastenings. Fig. 84 represents two
plates connected together by an ordinary trebleriveted butt
strap. The spacing of the rivets in the line of holes nearest
the butt is such that the joint can be caulked and made water
tight, and the alternate rivets are left out of the row of holes
farthest from the butt. Such a connection as this could con
ceivably break in five distinct ways
1. By the whole of the rivets on one side of the butt
shearing.
2. By the plate breaking through the line of holes, AA,
farthest from the butt.
3. By the butt strap breaking through the line of holes, BB,
nearest the butt.
4. By the plate breaking through the middle row of holes,
CC, and shearing the rivets in the line A A.
5. By the strap breaking through the middle row of holes,
CC, and shearing the rivets in the line BB.
2 3 6
Theoretical Naval Architecture.
It is impossible to make such a connection as this equal to
the strength of the unpunched plate, because, although we might
i
*JD.B
ffa t
ipi
f:
,a i<
>!
i
4 >
t
I . {
,
fT(
]
r ,
1ft
4 H
"!M
, ^ !<

; i i<
if
i.,M
r
.
/ if! %?
>4*
!
( Li
* ii
)i
f,
j<
^
t
j^
* 4 M
i h
* *
1
[i ii
i4l <
1
'f
i '[
tr
A.Ci
I.
i
,_
FIG. 84.
put in a larger number of rivets and thicken up the butt stiap,
FIG. 85.
there would still remain the line of weakness of the plate
through the line of holes, AA, farthest from the butt.
Calculation of Weights, etc. 237
The most efficient form of strap to connect two plates
together would be as shown in Fig. 85, of diamond shape.
Here the plate is only weakened to the extent of one rivethole.
Such an efficient connection as this is not required in ship con
struction, because in all the plating we have to deal with, such
as stringers and outer bottomplating, the plate is necessarily
weakened by the holes required for its connection to the
beam or frame, and it is unnecessary to make the connection
stronger than the plate is at a line of holes for connecting it to
the beam or frame. In calculating the strength of a butt con
nection, therefore, we take as the standard strength the strength
through the line of holes at a beam or frame, and we so
arrange the butt strap that the strength by any of the modes
of fracture will at least equal this standard strength.
Experimental Data. Before we can proceed to calcu
late the strength of these butt connections, we must have some
experimental data as to the tensile strength of plates and the
shearing strength of rivets. The results of a series of experi
ments were given by Mr. J. G. Wildish at the Institution of
Naval Architects in 1885, and the following are some of the
results given :
SHEARING STRENGTH OF RIVETS IN TONS.
(Pan heads and countersunk points.)
Single shear. Double shear.
\ inch iron rivets in iron plates ... 10*0 18
3 ,, steel 84
\ inch steel ,, ,, ,, ii'S 21'2
i ,, ,, I525
i'o ,, ,, ,, ,, 2025
It will be noticed that the shearing strength of the steel
rivets of varying sizes is very nearly proportionate to the sec
tional area of the rivets. Taking the shearing strength of a
finch steel rivet to be 11*5 tons, the strength proportionate to
the area would be for a Jinch steel rivet, 15 '6 tons, and for a
iinch steel rivet 20*4 tons. Also, we see that the double shear
of a rivet is about 18 times the single shear.
The following results were given as the results of tests of
mild steel plates :
Theoretical Naval Architecture.
Unpunched 28J tons per square inch.
Holes punched 22 ,, ,,
or a depreciation of 22 per cent.
Holes drilled 29^ tons per square inch.
Holes punc /W small, and the hole then \
countersunk /
29
The following give the strength of the material of the plates
after being connected together by a butt strap :
249 tons per square inch.
Holes punched small and then countersunk, I
the rivets being panhead, with countersunk > 28*9 ,,
points
It appears, from the above results, that if a plate has the
holes drilled or has them punched and countersunk in the
ordinary way as for flush riveting, the strength of the material
is fully maintained. Also that, although punching holes in a
plate reduces the strength from 28^ to 22 tons per square inch,
a reduction of 22 per cent., yet when connected by a butt strap,
and riveted up, the strength rises to 249 tons per square inch,
which is only 12 per cent, weaker than the unpunched plate,
the process of riveting strengthening the plate.
The following table giving the results of more recent experi
ments, go to confirm the above figures :
TABLE OF BREAKING STRESS IN TONS PER SQUARE INCH (OF AREA
OF PLATING BEFORE TESTING).
Nature of test.
Mild
steel.
High tensile
steel.
(2) 28*0
Elongation % in 8 in
(15) 25 %
(8) 2Q'6
(2) 298 %
(2) 30'6
(5) 247 %
(8) 3i '6
Plate punched not countersunk, unriveted
ii n ii riveted .
,, ,, countersunk and riveted .
(16} 265
(33) 29*6
(16) 296
22'I
247
282
(6) 267
(12} 252
(12) 290 .
Single 5 in. rivet, counter* Mnd , .
sunk point, tons per H T rfvet
Double.shear in. rivet, \ M; , , ,
countersunk point, tons 1 Mild steel rivet
per rivet
O) i4'8
(2) 217
(2) 39'7
*5'7
289
(2) 210
Figures in brackets ( ) denote the number of tests of which the result given is the mean.
In an ordinary buttstrap, with the holes spaced closely
Calculation of Weights, etc. 239
together in order to obtain a watertight pitch for the rivets, it
is found that the punching distresses the material in the neigh
bourhood of the holes, and the strength is materially reduced,
as we have seen above, but after riveting the strength is to
some extent restored. It was formerly the practice to anneal
butt straps of steel plating, but this practice is now discontinued
in both Admiralty and Lloyd's practice.
In our calculations of the strength of butt straps, we assume
that the strength of the material between the rivetholes is the
same as the strength of the material of the unpunched plate.
Again, the plating, in the cases we have to deal with, has
the riveting flush on the outside, and the holes are made with
a countersink for this purpose. Here also we can assume that
the strength of the material is the same as the strength of the
material of the unpunched plate.
The specified tests for the tensile strength of mild steel
plates are as follows :
For ships built for the British Admiralty, not less than 26
and not more than 30 tons per square inch of section.
For ships built to the rules of Lloyd's Register, not less
than 28 and not more than 32 tons per square inch of
section.
The plates tested above showed a tensile strength of about
28 tons per square inch, or nearly midway between the limits
laid down by the British Admiralty. It seems reasonable,
therefore, in calculating the ultimate strength of riveted joints,
to take as the strength of the material the minimum strength to
which it has to be tested. Thus, in a ship built for the British
Admiralty, we can use 26 tons as the strength per square inch
of section, and in a ship built under Lloyd's rules, we can use
28 tons per square inch of section.
The following two examples will illustrate the methods
adopted in calculating the strength of butt fastenings :
i. A steel stringer plate is 48 inches broad and f 5 inch thick. Sketch
the fastenings in a beam and at a butt, and show by calculations that the
butt connection is a good one.
(S. and A. xam., 1897.)
For a fginch plate we shall require f inch rivets, and setting these out
240 Theoretical Naval Architecture.
at the beam, we require 9 rivets, as shown in Fig. 84. The effective breadth
of the plate through this line of holes is therefore
48 9(i) = 41* inches
and the strength is
41^ X ^ X 26 = 470 tons
and this is the standard strength that we have to aim at in designing the
butt strap.
(1) As regards the number of rivets. The shearing strength of a $inch
rivet being 1 1 '5 tons, the number of rivets necessary to equal the standard
strength of 470 tons is
470
^^ = 40*8, say 41 rivets
If we set out the rivets in the strap as shown in Fig. 84, leaving the
alternate rivets out in the line AA, it will be found that exactly 41 rivets
are obtained, with a fourdiameter pitch. So that, as regards the number
of rivets, the butt connection is a good one.
(2) The strength of the plate in the line A A is the same as at the beam,
the same number of rivetholes being punched in each case.
(3) If the strap is / s inch thick, the strength of the strap in the line BB
is given by
(48 i6(5)} x / fi X 26 = 410 tons
This is not sufficient, and the strap must be thickened up. If made \ inch
thick, the strength is
(48  i6(fl} X i X 26 = 468
which is very nearly equal to the standard strength of 4/0 tons.
(4) The shear of the 9 rivets in the line A A is 1035 tons so tnat tn e
strength of the plate through the line of holes CC and the shear of the
rivets in the line AA are
410 + 1035 = 5 T 3'5 tons
(5) Similarly, the strength of the strap through the line CC and the
shear of the rivets in the line BB are
468 + 184 = 652 tons
The ultimate strengths of the butt connection in the five different ways it
might break are therefore 47 1, 470, 468, 513^, 652 tons respectively, and
thus the standard strength of 470 tons is maintained for all practical
purposes, and consequently ihe butt connection is a good one.
2. If it were required to so join two plates as to make the strength at
the butt as nearly as possible equal to that of the unpierced plates, what
kind of butt strap would you adopt ?
Supposing the plates to be of mild steel 36 inches wide and inch thick,
give the diameter, disposition, and pitch of rivets necessary in the strap.
(S. and A. Exam., 1895.)
The first part of this question has been already dealt with on p. 237.
To lessen the number of rivets, it is best to use a double butt strap, as
Fig. 85, so as to get a double shear of the rivets. Each of the butt straps
should be slightly thicker than the halfthickness of the plate, say T 5 S inch.
The standard strength to work up to is that of the plate through the
single rivethole at the corner of the strap. inch rivets being used, the
standard strength is
(36  I) X X 26 = 457 tons
Calculation of Weights, etc. 241
The single shear of a inch rivet is 15! tons, and the double shear may be
taken as
1525 x i'8 = 27^ tons
and consequently the least number of rivets required each side of the
butt is
452 J66, say 17 rivets
The strength of the plate along the slanting row of holes furthest from
the butt must be looked into. The rivets here are made with a watertight
pitch, say from 4 to 4^ diameters. If we set out the holes for a strap 2 feet
wide, it will be found that the strength is below the standard. A strap
3 feet wide will, however, give a strength through this line of about 465
tons, which is very near the required 457 tons. There are 13 rivets along
the edge of the strap, and the inside may be filled in as shown, giving a
total number, of rivets, each side of the butt, of 19.
For the strength of an assemblage of plating like the outer
bottom, we must take the strakes as assisting one another. If
two passing strakes are assumed, then we can take a butt with
a through strake each side, and see how the strengths by
various methods of fracture compare with the standard strength
at a frame.
For the strength of plating at a watertight bulkhead, the
bulkhead liner is associated with the outside strake and one
half the adjacent inside strakes, and the strength should be
brought up to that at an ordinary frame.
Professor Hovgaard, in " Structural Design of Warships,"
deals very exhaustively with the above. In particular he
allows for the reduction of area caused by countersinking and
the slightly greater diameter of hole in the plate as compared
with the nominal diameter of the rivet.
Strength of Davits. The size of davits for merchant
vessels are usually fixed by the rules of a Registration Society.
The following is the rule adopted by Lloyd's Register, viz. :
For boats and davits of ordinary proportions the diameter
in inches is onefifth of the length of the boat in feet.
Where the height and spread of davits or dimensions of
boats are not of ordinary proportions the diameter of davit
in inches is found by the formula
= VLXBXD/H X
V 40 \3 /
242 Theoretical Naval Architecture.
Where L.B.D. are the dimensions of the boat, H is the
height and R the outreach from the point of support in feet.
The rule of the British Corporation is of the same form
but slightly different, viz. :
It is usual in H.M. service to test a davit to twice its
working load, and this test load is used to calculate the dimen
sions. If W be the load in tons, r the outreach in inches, then
the bending moment is W x r inch tons, and we apply the
formula
= y to find the diameter d, (y = J . d).
/w7>
'
Example. A boat weighing 2 tons is carried in davits with an out
reach of 6 ft. 6 in. Determine the diameter of davit, allowing a stress on
the material of 5 tons per square inch.
The moment WXr=2X78 = i56 inch tons, and the diameter is
given by
Example. A boat weighing I ton is carried in davits with an out
reach of 6 ft. 6 in. Determine the diameter of davit, allowing a stress on
the material of 5 tons per square inch.
Moment = 78 inch tons
This davit was made sJ in. diameter, and 3$ in. at head and heel, the
bow of davit being flattened out to an oval shape 5^ in. x 4f in.
Example. A davit with outreach of 7 ft. 6 in. is tested to 3 tons.
Find the maximum compressive and tensile tresses, the diameter of davit
being 7 in.
The max. BM is
3 x 90 = 270 inch tons
I of crosssection = V s . ^ . 49 = ir . 49 * 49
4 64
y = 3'S
. * M 270 X 35 X 64
'^ = T'^ = ^9x49 = 8topg **"
Calculation cf Weights, etc. 243
There is an additional compressive force due to the weight, viz.
3 i  = = *o8 tons sq. in. The tensile stress will be dimin
4 49' 7r
ished by this amount.
/. Compressive stress = 8'o8 tons sq. in.
Tensile stress = 7*92 tons sq. in.
Davit Diagram. The following method of drawing once
for all a davit diagram has beea brought to the Author's notice
by J. J. KingSalter, Esq., R.C.N.C. Its use is very simple
and obviates the necessity of calculating davit and similar
diameters.
If units are taken in Ibs. and inches, say a weight of w Ibs.
and an overhang of r inches, then if a working stress is
assumed of 4*5 tons per square inch, the diameter of a davit
can be expressed in the simple form
, // X r t .
* m \* ' ...... (i)
V 1000
This is seen to depend on the product w X r. It can readily
be shown that for a rightangled triangle, where a is the per
pendicular from the right angle t< ) the opposite side and b and
c are the divisions of that side by the perpendicular, then
a? = b X c\ or = *] b .c
In Fig. $5A, above the base t the side is set up a scale of
overhang in inches r> and below a scale of weight in Ibs. w.
Along the base is set off a scale of *Jw .r. Thus for w =
10,000 Ibs. and r  100, */w.r=iooo. This point must
subtend a right angle to the values of w and r taken, and this
will determine the scale to use along the base.
Now at various points along the base set up the corre
sponding value of d from the formula (i) above. Thus where
V w . r 1000, w . r = 1,000,000 and d = 10 " ; d = 5" at an
abscissa of 354, and so on. Through the spots thus obtained
a curve of diameters is drawn as shown.
The method of use is to employ a set square with the sides
forming the right angle passing through the values given by
the problem for w and r, with the right angle on the base line.
Theoretical Naval Architecture.
An ordinate from this point to the curve will give the diameter
required. Thus for w = 4000 Ibs., r = 70", d is found to be
6*6 in. If diameter is 8 in. say, and overhang is 80 in., the
load is 6400 Ibs., and
so on.
The diagram can be
drawn out on a large
scale and mounted for
general drawing office
use.
Pillars. Gordon's
Formula. The for
mula usually employed
to determine the strength
of pillars is that known
as Gordon's formula, as
follows :
W is the crippling
load, A the cross sec
tion, f the stress given
10,000 * in table, n is obtained
from I = n . A . > 2 , where
h is the least breadth,
5.000
c is a coefficient given in the table.
W
I \ 75
c.n.tf
n = ^ for a rectangular section and yg for circular section ; for
a circular hollow section n = .
Units are taken as tons and inches.
Material
/
c
tons sq. in.
Ends free.
One end fixed.
Both ends fixed.
Wrought iron .
Mild steel . .
16
3
9,000
9,OOO
18,000
18,000
36,000
36,000
Cast iron . .
35
1,600
3,200
6,400
Dry timber . .
3
750
1,500
3,ooo
Calculation of Weights, etc.
245
This formula is empirical. For a discussion regarding its
use the student is referred to such works as Lineham's
" Mechanical Engineering."
Example. A cargo derrick for a vessel is constructed of steel plating
5 ^5 in. thick and two T bars 5 in, X 3 in. X 5 8 in. The jib is 40 ft. long, and
the topping lift is led to a point on a mast 32 ft. above heel of derrick. The
/S
FIG. 858.
maximum load to be lifted being 16 tons ; calculate the approximate
diameter of the derrick if the maximum stress on the material is not to
exceed 4 tons per sq. in. (neglect effect of T bars).
(Honours B. of E. 1909.)
By setting out the triangle of forces CDE at the head of derrick the
thrust on derrick is found to be 20 tons, DE being parallel to CB (Fig.
In the above formula f 4, c
the unknown diameter.
9000, n J, A =  .d % d being
246 Theoretical Naval Architecture.
'* ~
"!*.</ ,.+
9000 x 
or o25^ 3 <f>^,&\$
By trial d is found to be 1 1 inches nearly.
Example. A wooden derrick 34 ft. long when tested to twice the
working load is found to be subject to a thrust of 4^ tons. Determine the
diameter, allowing a factor of safety of 6 when being tested.
26
TONS
FIG. 850.
In the above formula, W = 255 tons, /= 3, c  750, / = 34 X 12,
n T l s , d diameter.
Tr.a' (34 X 12)'
4 750 X T ' 8 X a
= 0092^
or cro92</ 4  d? = 3550
By trial d is found to be about 14^ in.
Calcidation of Weights, etc. 247
Example. A boat hoisting derrick 60 ft. long, estimated weight 6 tons,
is arranged as shown on figure herewith, the purchase being single through
sheaves A and D. The topping lift has the fixed part at E, and passes
through sheaves B and C. Determine the forces on blocks and ropes and
the thrust on derrick, when holding the test load of 26 tons. Determine
the diameter of the derrick if formed of in. steel plating, the T bars
forming edge strips being neglected, and a factor of safety of 5 being
assumed. (Fig. 850.)
oa is set down = 29 tons, i.e. 26 tons plus half weight of derrick, and
ab is drawn parallel to the purchase AD, and equal to 26 tons. Then ob
is the resultant force at the head of derrick due to the forces on the purchase
and the weight of derrick, be is then drawn parallel to the topping lift.
Then obc is a triangle of forces, giving for the force on topping lift
be 35 tons and the thrust on derrick oc = 59 tons. The force in the link
AB is therefore 35 tons, and the two parts of topping lift have each a
force of 17*5 tons. The force on the block C is found by drawing the
triangle of forces deg, de = eg=\T$ tons from which force on block
is dg 29 tons. Similarly the force on block D is found to be
38 tons.
The length of derrick from pin of sheave to the trunnion is 56*5 feet.
In Gordon's formula we have therefore
W = 59 X5 = 295 tons,/= 30, c 9000, n = J, d y the diameter, is the
unknown, A = . TT . </, / = 56*5 X 12.
We have therefore
30 X j x y x d
295 ~ ! , (565 x 12)*
9000 X 5 X d*
410
or I + ^  = O'l2d
or o'izd 3 d? 410
from which d is found by trial to be i8 ins. nearly.
It may be noted that the above derrick was actually made
20 in. diameter, of plating, 14 Ibs. per square foot, which allows
for the loss due to the rivets in butt strap.
Fig. 850 illustrates the case where electric winches are
employed for both topping lift and purchase, and the
greater speed of these winches renders more turns of rope
necessary.
The test load is 32 tons, being 'twice the weight of the
boat. This with the weight of the block A gives 324 tons,
which is taken by the three ropes supporting A, giving 108
tons to each. To find the force at the topping lift and on
the derrick, we draw the diagram of forces, shown on top
of figure, ab = 36 tons, i.e. 32 + 36 (half weight of der
rick) + 04 (weight of block A) ; be io'8 is drawn parallel
248
Theoretical Naval Architecture.
to the purchase, so that ac is the force at the head of the
derrick ; cd is drawn parallel to the topping lift, and ad
parallel to the derrick. Then cd is force on topping lift = 59
tons, and ad the thrust on derrick = 60 tons. The topping
lift is in four parts, giving 15 tons to each, and the block C has
59 tons. The block D has three parts, viz. 45 tons. The block
E has 15 tons along each of the ropes, and as shown by the
72 TONS
FIG. 850.
diagram, sustains a force of 26 tons. The block F has 10*8
tons along each of the ropes, and, as shown by the diagram,
sustains a force of 18 tons. The strength at the heel of the
derrick can be allowed for knowing the thrust to be 60 tons.
The test loads of the various parts can now be allowed for,
being usually about twice that due to the 32ton test load
applied.
Calculation of Weights, etc.
249
Fig 85 E shows an ordinary form of derrick for a cargo
vessel, the derrick being supported by a stump mast. The
purchase is taken round two single blocks A and B, and thence
DOUBLE
D
FIG. 858.
to the winch. The topping lift has a single block at C, a
double block at D, and a single block at E, and thence to the
winch. Taking a load of 5 tons the triangle of forces abd is
drawn ab ac 5, and the resultant force due to the load is
ad 9 tons, which is the force on the block A. de is drawn
parallel to the topping lift. Then ae = 14 tons is the thrust
250
Theoretical Naval Architecture.
on the derrick, and 67 tons is the force on block C. This
divided into 3 gives 2*2 tons in each portion of the topping
lift. The force on block D is obtained by drawing the
triangle of forces fgh. Similarly the forces on the blocks B
and E are obtained. The forces thus obtained give a basis
for estimating the strength of all the parts, including the
derrick and the stump mast.
Shaft Brackets. (By A. W. Johns, Esq., R.C.N.C.).
The length and diameter of the drum or barrel of a shaft
bracket are determined by the requirements of the engineer.
The inside diameter is arranged to take the shaft and its
bearings and bushes. The outside diameter is usually from
3 to 6 inches greater than the inside diameter, depending on
the size of shaft The inside surface is generally gulleted to
a depth of from i to ij inches. The length of barrel is
governed by the length required by the engineer's bearings
in the bracket (see Fig. 85 F).
The length of the arms or struts must necessarily depenc
on the position of the axis of the shaft at the bracket and the
shape of the ship in the vicinity. The section of the arms is
usually pearshaped (see Figs. 850 and 108 for examples)
with the blunt end forward. The dimensions of the section
must be governed by the straining action to which the
bracket is subjected. Formerly these dimensions appear
Calculations of Weights* etc. 251
to have been determined in a roughandready way from the
experience of the designer responsible. Knowing the dimen
sions in previous cases which on service had proved sufficiently
FIG. 850.
strong, he would vary these dimensions in a new ship accord
ing to the variation of the horsepower, or perhaps the size and
overhang of the tail shaft. Consequently it will be found that
ships of about the same size, horsepower and revolutions pro
duced under different designers, have entirely different dimen
sions (and weights) of shaft brackets.
At first sight a suitable basis of comparison for such dimen
sions appears difficult to obtain, but investigation proves that
the matter is a comparatively simple one, as is seen by what
follows :
With the centre of gravity of the revolving parts, viz. shaft
and propeller, in the axis of rotation, the straining actions
which may operate on a bracket are as follows, viz. :
1. Forces due to the weight of the propeller, shaft
and bracket. These are equivalent to a downward force
on the bracket, and a bending moment on it, equal to the
difference in the moments of weight on the forward and
after sides. Both the force and the bending moment may
be increased appreciably by the accelerative effect during
pitching.
Thus in a given ship 500 feet long pitching in a " single " period of
3 seconds, the maximum acceleration is 250 X , . X 0, where is angle of
pitching (see Chap. IX. on Rolling).
If 9 = 4, say J 4 in circular measure, acceleration = 20 in footsecond
units. This added to the acceleration due to gravity gives 52^2 or a
virtual weight of I '6 times the actual, and all the forces are increased in
this ratio.
2. Forces called into play when pitching or turning due to
the gyroscopic action of the propeller and shaft.
252
Theoretical Naval Architecture.
3. Forces caused by unequal pressure on the blades of the
propeller when the ship is turning. Here, owing to the trans
verse motion of the stern, the forces on the blades above the
horizontal will be different to those below.
If, however, the centre of gravity of the revolving weights
is not in the axis of rotation, there will be in addition to the
above forces a centrifugal force operating which will tend to
bend the shaft where it enters the strut, and will also tend to
bend and twist the bracket. At high revolutions, which is the
case in turbine machinery, heavy straining actions are set up
if a propeller blade is broken or lost.
The following are approximate values of the various forces
and moments considered above, worked out for the case of
a large cruiser.
Force on bracket.
Moment on bracket.
(i) Due to weight of propeller,
etc
30 tons
90 foot tons
(la) Due to weight of propeller,
etc., when pitching . .
(2) Gyroscopic action . . .
(3) Turning at full speed with
full rudder angle . . .
(4) Centrifugal action due to
the loss of a propeller
blade at full revolutions
48
12
100
M4
5
80
5 '
It will be seen that the last case produces by far the
heaviest straining effect. In addition to straining the bracket,
however, the shaft also will be strained, the maximum stress
occurring at the section immediately at the after end of the
barrel of the bracket. For good design the bracket should
be stronger than the shaft, for then the shaft would break at
the after end of the boss of shaft bracket, whereas if the
bracket were weaker than the shaft the former would first
break and the shaft losing its after support would then bend
or break with perhaps disastrous effect on the ship. A basis
of calculation is therefore obtained by considering the strength
of the shaft and making the shaft bracket somewhat stronger.
If T is the force at the propeller (Fig. 85 F), producing
Calculation of Weights, etc. 253
a bending moment T . a on the shaft which will just bring the
material of the shaft to its full working strength, this force T
must be employed in determining the dimensions of the arms
of the bracket and also in determining the number, size
and spacing of the rivets connecting the bracket to the hull.
T acting at propeller is equivalent to
1. A parallel force T acting directly on the bracket, and
2. A moment k . T . m on the bracket, where k has an
average value of about 065.
If T is caused by centrifugal action, then as the shaft
revolves it is always being bent in the same way, but the
bracket being fixed the force and moment on it are constantly
altering in direction. 1 Bending alone occurs when the line of
action of T lies in a plane passing through the axis of shaft
and bisecting the angle between the arms. Twisting occurs
when the line of action of T is perpendicular to that plane.
For other directions of the line of action of T combined bend
ing and twisting occur.
Generally bending alone produces the greatest stress on
the shaft arms, and this produces a stress given by
.T. m. y.cos
p ~ ~ir
where I is the moment of inertia of a right section of the arm
about an axis through the geometrical centre and
perpendicular to the longer dimension of the section.
y is the distance of the most strained layer from this axis ;
and 6 is half the angle between the arms.
For ordinary pearshaped sections,^ = 0*55 R and I = gV'R 8 . r,
where R and r are the longer and shorter dimensions of the
section of the arm.
Taking, say, 6 tons as the maximum working strength of
the shaft, the force T necessary to strain the shaft to this limit
can readily be found when a the overhang and D and d the
external and internal diameters of the shaft are known. Sup
posing the shaft bracket is of cast steel and taking 4^ tons as
the working strength (5 tons is really allowed, but \ ton is
1 The loss of a propeller blade is soon evident, for the ship will vibrate
violently if the revolutions of the engine approach the full number.
254
Theoretical Naval Architecture.
allowed for the force T acting directly on the bracket), the
following relation is obtained
R 2 .r=o63X ^ x^Xcosfl . . (i)
All dimensions being in inches.
Usually 6 = about 45 and we then have
D 4 
m
R 2 . r = 044 X ^ X 
If, however, 9 is small, we have approximately
D 4 d* m
R 2 . r = 063 X
D
<*)
(3)
FIG.
As stated above, the stress produced on the bracket by
bending is usually greater than that
produced by twisting, but in the case
where the angle between the arms is
small the stress due to twisting should
also be investigated. This can be
done as follows :
Taking, as in the figure 85 H, was
the distance between the centres of
the arms, A the area of each arm,
q the stress in the arm, the moment
resisting twisting is given by q . A . n.
This must equal k . T . m, and hence the stress due to twisting
becomes
k .T. m . A
q= A n (A =
q is a shearing stress and the equation
shows that if A is constant q increases as
n decreases. Close to the barrel q is a
maximum, and is a minimum where the
arms enter the hull. For q to be constant
A should vary inversely as n. Usually,
however, A is kept constant and the arms
are run tangential, as in Fig. 85;, instead of radial to the
barrel. This has the effect of increasing n near the barrel and
diminishing q.
Equations (i), (2), and (3) above can be used to determine
FIG. 85;.
Calculation of Weights, etc.
255
the value of R 7 r, where the bracket is of cast steel. If the
bracket is of other material the working strength of the latter
must be substituted for the 4^ tons used above.
It will be noticed that economy of material is obtained by
making the ratio R r r as large as possible, for since the
square of R enters into the relation it has far more influence
than r, which appears in the first power only. Thus if R a . r
= 8000 and R = $r, then R = 29 in. and r = Q in. and A
= 212. Whereas if R = 6r, R = 36 in., and r = 6 in., and
A = 162 or a saving in weight of about 25 per cent. There
is also an appreciable reduction in resistance. Taylor's ex
periments with shaft brackets show that resistance in Ibs. per
foot length of shaft bracket arm is given by
F = ^(A + 4o)V a
IOOO V
where V = speed in knots
A = area of section in square inches, for values between
40 and 175 sq. inches.
c = a constant depending on the ratio .
VALUES OF c.
Ratio Jl
r
3
4
5
6
7
8
9
10
ii
12
Value of c
188
132
107
094
086
080
076
074
072
071
Further than this there is little doubt that with bracket
arms whose ratio R r r is small, at high speeds a large amount
of dead water trails behind the arms reaching to the propeller
disc and causing vibration and loss of propeller efficiency as
the blades of the propeller enter and leave the dead water.
By increasing the ratio R 4 r these effects are diminished.
Fig. 85 E gives the section of the arm of shaft bracket of a
recent ship of large power and great speed.
Finally it is interesting to compare the dimensions given
by formula (2) above with those adopted in practice in parti
cular cases as the result of experience. The ratio R f r has
254
Theoretical Naval Architecture.
allowed for the force T acting directly on the bracket), the
following relation is obtained
= 063 X
D 4 
D
m
x Xcos0
(0
All dimensions being in inches.
Usually 6 = about 45 and we then have
D 4 d* m
r = 044 :
R 2
1)
If, however, 9 is small, we have approximately
D* d* m
13*. r= 063 X ^ X 
(3)
D a
As stated above, the stress produced on the bracket by
bending is usually greater than that
produced by twisting, but in the case
where the angle between the arms is
small the stress due to twisting should
also be investigated. This can be
done as follows :
Taking, as in the figure 85 H, was
the distance between the centres of
the arms, A the area of each arm,
q the stress in the arm, the moment
resisting twisting is given by q . A . n.
This must equal k . T . m, and hence the stress due to twisting
becomes
.T. m
FIG.
q is a shearing stress and the equation
shows that if A is constant q increases as
n decreases. Close to the barrel q is a
maximum, and is a minimum where the
arms enter the hull. For q to be constant
A should vary inversely as ;/. Usually,
however, A is kept constant and the arms
are run tangential, as in Fig. 85;, instead of radial to the
barrel. This has the effect of increasing n near the barrel and
diminishing q.
Equations (i), (2), and (3) above can be used to determine
FIG. 8sj.
Calculation of Weights, etc.
255
the value of R 7 r, where the bracket is of cast steel. If the
bracket is of other material the working strength of the latter
must be substituted for the 4^ tons used above.
It will be noticed that economy of material is obtained by
making the ratio R 4 r as large as possible, for since the
square of R enters into the relation it has far more influence
than r, which appears in the first power only. Thus if R a . r
8000 and R = 3r, then R = 29 in. and r = Q in. and A
= 212. Whereas if R = 6r, R = 36 in., and r = 6 in., and
A = 162 or a saving in weight of about 25 per cent. There
is also an appreciable reduction in resistance. Taylor's ex
periments with shaft brackets show that resistance in Ibs. per
foot length of shaft bracket arm is given by
F = (A + 40) V a
IOOO
where V = speed in knots
A = area of section in square inches, for values between
40 and 175 sq. inches.
n
c a constant depending on the ratio .
VALUES OF c.
Ratio IL
r
3
4
5
6
7
8
9
10
ii
12
Value of c
188
132
107
094
086
080
076
074
072
071
Further than this there is little doubt that with bracket
arms whose ratio R 4 r is small, at high speeds a large amount
of dead water trails behind the arms reaching to the propeller
disc and causing vibration and loss of propeller efficiency as
the blades of the propeller enter and leave the dead water.
By increasing the ratio R f r these effects are diminished.
Fig. 85 E gives the section of the arm of shaft bracket of a
recent ship of large power and great speed.
Finally it is interesting to compare the dimensions given
by formula (2) above with those adopted in practice in parti
cular cases as the result of experience. The ratio R f r has
2 5 6
Theoretical Naval Architecture.
been kept the same in the calculation as actually adopted.
All dimensions are in inches.
Actually fitted.
Calculated.
R
r
R
r
Cruiser .
7f
37
12
14
4
133
ri
9
21
16
6
168
63
16
10
10
7
5
62
78
27
27
33
20
11
6
6
8
2i'5
24
301
!
86
21
ii
76
30
32
10
312
98
Battleship
Destroyer
20
84
4*
72
36
18
27
13
2
30
2'2
The relations in d), (2), and (3) given above apply strictly
to the usual pearshaped section of arm, but the method indi
cated can be applied to any particular case and the dimensions
calculated.
EXAMPLES TO CHAPTER VI.
1. The area of the outer bottom plating of a ship, over which the
plating is worked 25 Ibs. per square foot, is 23,904 square feet, lapped
edges and butt straps, both doubleriveted. Estimate the difference in
weight due to working the plating with averagesized plates 20' X 4^', or
with the average size 12' x 3'.
Ans. About 20 tons.
2. Steel angle bars 3^" X 3" are specified to be 8 Ibs. per lineal foot
instead of ^ inch thick. Determine the saving of weight per 100 lineal
feet.
Ans. 52 Ibs.
3. Determine the weight per lineal foot of a steel T~bar 5" x 4" X J".
Ans. 1445 lbs
4. For a given purpose, angle bars of iron 5" X 3" X T 8 8 " or of steel
5" X 3" X 2 V' can be used. Find the saving of weight per 100 feet if steel
is adopted.
Ans. 95 Ibs.
5. A mast 96 feet in length, if made of iron, is at its greatest diameter,
viz. 32 inches, ^ inch thick, and has three angle stiffeners, 5" X 3" x ^".
For the same diameter, if made of steel, the thickness is 55 inch, with three
angle stiffeners 5" X 3" X 5 9 ". Estimate the difference in weight.
Ans. About I ton.
6. At a given section of a ship the following is the form : The lengths of
ordinates 3 feet apart are 19*6, 1885, r 7'8, 16*4, 145, 1 1 '8, 735, and
I'D feet respectively. Estimate the vertical position of the centre of
gravity of the curve forming the section, supposing it is required to find the
vertical position of the centre of gravity of the bottom plating of uniform
thickness.
Ans. About 12\ feet from the top.
Calculation of Weights, etc. 257
7. The halfgirths of the inner bottom of a vessel at intervals of 51 feet
are 26*6, 29*8, 320, 32 '8, and 31*2 feet respectively, and the centres of
gravity of these halfgirths are i8'6, 20*6, 21 '2, 20*0, 174 feet respectively
below the L.W.L. Determine the area of the inner bottom and the
position of its centre of gravity both longitudinally and vertically. If the
plating is of 15 Ibs. to the square foot, what would be the weight, allowing
14^ per cent, for butts, laps, and rivetheads.
Ans. 12,655 square feet ; 105 feet from mer end, 2O feet below the
L.W.L. ; 97 tons.
8. The whole ordinates of the boundary of a ship's deck are "S'5, 24,
29, 32, 33'5, 33'5, 33'5> 32, 3 2 7, and 65 feet respectively, and the
common interval between them is 21 feet.
The deck, with the exception of 350 square feet, is covered with \ inch
steel plating worked flush jointed, with single riveted edges and butts.
Find the weight of the plating, including straps and fastenings.
Ans. 45 tons.
9. A teak deck, 2^ inches thick, is supported on beams spaced 4 feet
apart, and weighing 15 pounds per foot run. Calculate the weight of a
middleline portion of this deck (including fastenings and beams) 24 feet
long and 10 feet wide. Ans. I '65 tons nearly.
10. Taking the net weight of outer bottom plating of a vessel as 1000
tons, estimate the saving of weight if the average size of plates is 20 feet
by 5 feet as against 18 feet by 3! feet. (Buttstraps double riveted, lapped
edges double riveted, inch rivets.) Ans. 43 tons about.
n. A longitudinal W.T. bulkhead is bounded at its upper edge by a
level deck (having 9inch beams, 4 feet apart) and at its lower edge by the
inner bottom. The depths of the bulkhead at ordinates 61 feet apart are,
commencing from forward, g'o, 16*7, 19*3, I5'4> 9'5 feet respectively.
The plating of the bulkhend is 15 Ibs. per square foot worked vertically,
single riveted, and the stiffening consists of Z bars of 12 Ibs. per foot
spaced 4 feet apart with intermediate angles of 7 Ibs. per foot. There is
a single boundary bar of 8'5 Ibs. per foot.
Calculate (i) the weight of the bulkhead.
(2) the distance of C.G. from forward end.
(3) the distance of C.G. below the deck.
Ans. (i) 39 tons ; (2) 1205 feet ; (3) 8 feet 
12. The half ordinates of upper deck of a ship 360 feet long are (i) o ;
(2) 9'4; (3) l6 ' 2 ; (5) 244; (7) 288; (9) 312; (ii) 324; (13) 322;
(15) 315 ; (17) 296 ; (19) 24 6 ; (20) 20'i ; (21) 138. Over the midship
portion (7) to (15) the beams are 24^ Ibs. per foot, 4 feet apart, and the
plating is 20 Ibs. per square foot, with singleriveted edgestrips and
doubleriveted buttstraps. At the ends the beams are 24^ Ibs., 3 feet
apart, completely covered with plating 10 Ibs. per square foot, lapped and
singleriveted. The boundary bar is 3 inches by 3! inches of 8 Ibs. per
foot, and the deck is completely planked with 3inch teak. Find total
weight, neglecting hatches, etc.
Ans. 325 tons about.
CHAPTER VII.
STRAINS EXPERIENCED BY SHIPS CURVES OF LOADS,
SHEARING FORCE, AND BENDING MOMENT
EQUIVALENT GIRDER "SMITH" CORRECTION
TROCHOIDAL WAVE.
Strains experienced by Ships. The strains to which ships
are subjected may be divided into two classes, viz.
1. Structural strains, i.e. strains which affect the structure
of the ship considered as a whole.
2. Local strains, i.e. strains which affect particular portions
of the ship.
1. Structural Strains. These may be classified as
follows :
(a) Strains tending to cause the ship to bend in a foreand
aft direction.
(b) Strains tending to change the transverse form of the
ship.
(c) Strains due to the propulsion of the vessel, either by
steam or sails.
2. Local Strains. These may be classified as follows :
(a) Panting strains.
(b) Strains due to heavy local weights, as masts, engines,
armour, guns, etc.
(c) Strains caused by the thrust of the propellers.
(d) Strains caused by the attachment of rigging.
((?) Strains due to grounding.
We will now deal with some of these various strains to
which a ship may be subjected in a little more detail.
Longitudinal Bending Strains. A ship may be regarded as
a large beam or girder, subject to bending in a foreandaft
direction. The support of the buoyancy and the distribution
of weight vary considerably along the length of a ship, even
Strains experienced by Ships, etc.
259
when floating in still water. Take a ship and imagine she is
cut by a number of transverse sections, as in Fig. 86. Each
of the portions has its weight, and each has an upward support
of buoyancy. But in some of the portions the weight exceeds
the buoyancy, and in others the buoyancy exceeds the weight.
The total buoyancy of all the sections must, of course, equal the
total weight. Now imagine that there is a watertight bulkhead
at each end of each of these portions, and the ship is actually
cut at these sections. Then the end portions (i) and (5) have
considerable weight but small displacement, and consequently
they would sink deeper in the water if left to themselves. 1 In
i
FUG. 86.
the portions (2) and (4), on the other hand, the buoyancy might
exceed the weight (suppose these are the foreandaft holds, and
the ship is light), and if left to themselves they would rise. The
midship portion (3) has a large amount of buoyancy, but also
a large weight of engines and boilers, and this portion might
very well have to sink a small amount if left to itself. In any
actual ship, of course, it is a matter of calculation to find how
the weight and buoyancy vary throughout the length. This
case is somewhat analogous to the case of a beam supported
and loaded as shown in Fig. 87. At each point along the
beam there is a tendency to bend, caused by the way the
beam is loaded and supported, and the beam must be made
1 Strictly speaking, each portion would change trim if left to itself, but
we suppose that the various portions are attached, but free to move in a
vertical direction.
260
Theoretical Naval Architecture.
sufficiently strong to withstand this bending tendency. In the
same way, the ship must be constructed in such a manner as
to resist effectually the bending strains that are brought to
bear upon the structure.
When a vessel passes out of still water and encounters
FIG. 87.
waves at sea, the strains to which she is subjected must differ
very much from those we have been considering above.
Suppose the ship to be end on to a series of waves having
lengths from crest to crest or from trough to trough equal to
the length of the ship. We will take the two extremes.
(1) The ship is supposed to have the crest of the wave
amidships.
(2) The ship is supposed to have the trough of the wave
amidships.
HOGGING
ACROSS WAVE TROUGH
FIGS. 88, 89.
(i) This is indicated in Fig. 88. At this instant there
is an excess of weight at the ends, and an excess of
buoyancy amidships. The ship may be roughly compared
to a beam supported at the middle, with weights at the end.
Strains experienced by Ships, etc.
261
as in Fig. 90. The consequence is that there is a tendency
for the ends to droop relatively to the middle. This is termed
(2) This is indicated in Fig. 89. At this instant there
is an excess of weight amidships, and an excess of buoy
ancy at the ends, and the ship may be roughly compared
to a beam supported at the ends and loaded in the middle,
as Fig. 91. The consequence is, there is a tendency for the
middle to droop relatively to the ends. This is termed
sagging.
FIG. oo.
We have seen above how the ship may be compared to a
beam, and in order to understand how the material should be
disposed in order best to withstand the bending strains, we
will consider briefly some points in connection with ordinary
beams.
FIG. 91.
Take a beam supported at the ends and loaded at the
middle. It will bend as shown exaggerated in Fig. 92. The
resistance the beam will offer to bending will depend on
the form of the section of the beam. Take a beam having
a sectional area of 16 square inches. We can dispose the
material in many different ways. Take the following :
262
Theoretical Naval Architecture.
(a) 8 inches wide, 2 inches deep (a, Fig. 93).
(<) 4 inches wide, 4 inches deep (, Fig. 93).
(c) 2 inches wide, 8 inches deep (<r, Fig. 93).
(d) 8 inches deep, with top and bottom flanges 5 inches
wide and i inch thick (d, Fig. 93).
FIG. 92.
Then the resistances of these various sections to bending
compare as follows :
If (a) is taken as i, then (b) is 2, (c) is 4, and (d)
is6f.
We thus see that we can make the beam stronger to resist
bending by disposing the material far away from the centre.
FIG. 93.
The beam (d) has 6f times the strength of (a) against bending,
although it has precisely the same sectional area. A line
drawn transversely through the centre of gravity of the section
of a beam is termed the neutral axis.
In the British Standard Sections it will be found that for
Z bars and channel bars the flanges are distinctly thicker
than the web.
These principles apply equally to the case of a ship, and
we thus see that to resist bending strains the material of the
Strains experienced by Ships, etc.
263
structure
axis. 1
should be disposed far away from the neutral
For hogging strains, the upper portions of the vessel are
in tension and the lower portions are in compression. For
sagging strains, the upper portions are in compression and the
lower portions are in tension. Thus the portions of the struc
ture that are useful in resisting these hogging and sagging strains
are the upper and main decks and stringers, sheer strake and
plating below, plating at and below the bilge, both of the inner
and outer bottom, keel, keelsons, and longitudinal framing.
Strains tending to change the Transverse Form of the Ship.
Strains of this character are set up in a ship rolling heavily.
Take a square framework joined at the corners, and imagine
it to be rapidly moved backwards and forwards as a ship does
when she rolls. The framework will not break, but will distort,
as shown in Fig. 94. There is a tendency to distort in a similar
FIG.
way in a ship rolling heavily, and the connections of the beams
to the sides, and the transverse structure of the ship, must be
made sufficiently strong to prevent any of this racking taking
place. Transverse bulkheads are valuable in resisting the
tendency to change the transverse form.
A. ship, when docked, especially if she has on board heavy
weights, as armour or coals, is subjected to severe strains
tending to change the transverse form. If the ship is supported
1 There are other strains, viz. shearing strains, which are of importance
(see later).
266
Theoretical Naval Architecture.
points along the beam, we shall be able to draw a line through all
the spots as amb, which has a maximum ordinate at the centre
o of J . W . a. This line will give the bending moment at any
point along the beam.
Or take the case of a beam supported at the ends and
loaded uniformly with the weight w per foot run, the total weight
w.o/
being, therefore, 2 . w . a, as Fig.96. The support at each end
is w . a. The bending moment at any point K, distance x from
the end, is M k = w . a . x J . w . oc*. When x a this bend
ing moment is therefore J . w . a 2 . If a number of spots be
thus obtained throughout the length of the beam, we can draw
a curve as amb, any ordinate of which will give the bending
moment at that point of the beam.
FIG. 97.
Take now the case of a beam supported at one end and
loaded uniformly. The load can be graphically represented by
a rectangle oabc> ab = w (Fig. 97). At any point K, calculate
Strains experienced by Ships, etc.
267
the area of the rectangle ob on one side of k, i.e. w . x, and set off
as an ordinate kf = w .x. This is what is termed the " shearing
force " at K, or the tendency the two consecutive sections of
the beam at K have to slide over one another. Doing this
all along the beam, we should obtain the line afg> the maximum
ordinate of which og = w .1.
The area of the figure akf= .w.x*, and the bending momen*
at K also = % . w . x?. So that to determine the bending momem
at any point, we find the area of the curve of shearing force up
to that point. 1 In this way the curve of bending moment
amm is constructed, having a maximum ordinate om of \ . wl 2 .
This method of determining the bending moment at any point
from the curve of shearing force is of no value in this particular
case, but is of assistance when dealing with more complicated
cases cf loading.
Take, for example, a beam similar to the above, but loaded
unevenly along its length, such that the intensity of the load
FIG. 98.
at any point is given by the ordinate of the curve //, which we
may term a " curve of loads'' as Fig. 98. Take any point K
and determine the area beneath the curve of loads from the
point k to the end of the beam. This will give the shearing
force at K. Doing this all along the beam, we can draw the
curve of shearing force aff. The area under this curve between
1 For the proof of this in any general case, see any standard work on
*' Applied Mechanics," as Cotterill, chap. iii.
268
Theoretical Naval Architecture.
k and the end of the beam gives the bending moment at K,
and in this manner the curve of bending moment amm Can
be obtained.
Turning now to the case of a ship floating in still water.
There will be a certain distribution of the weight and also of
the buoyancy. The total weight must, of course, be equal to
the total buoyancy, and also the foreandaft position of the
centre of gravity of the weight must be in the same athwartship
section as the centre of buoyancy. But although this is so,
the distribution of the weight and buoyancy along the ship
must vary from section to section.
FIG. 99
Take the case of a vessel floating in still water in which
the buoyancy exceeds the weight amidships, and the weight
exceeds the buoyancy at the ends. Let BB in Fig. 99 be the
" curve of buoyancy." The area under this curve will give
the displacement of the vessel, and the foreandaft position of
the centre of gravity of this area is the same as the foreandaft
position of the centre of buoyancy.
Also let WW be the " curve of weight." This curve is
constructed by taking all the weights between two sections and
setting up a mean ordinate to represent the total weight between
the sections. This done throughout the length gives a number
of spots through which a curve may be drawn as nearly as
Strains experienced by Ships, etc. 269
possible. This curve should be adjusted as necessary to fulfil
the conditions stated above, viz. that the area under it shall
equal the area of the curve of buoyancy, and the foreandaft
position of the centre of gravity of the area under it shall be
in the same section as that of the curve of buoyancy.
The difference at any point between the ordinates of the
two curves WW and BB will give the difference between the
weight and the buoyancy at that point. Where the curves
cross at A and B, the weight and buoyancy are equal, and the
sections at these points are said to be " waterborne"
Now set off ordinates all along, giving the intercept between
the curves WW and BB. Set below the baseline where the
weight exceeds the buoyancy, and above the baseline where
the buoyancy exceeds the weight. In this way we obtain the
curve LLL which is the " curve of loads" At the sections
where the curve of loads crosses the baseline the ship is
waterborne. We now obtain the "curve of shearing force"
FFF from the curve of loads by finding the area under the
curve of loads, as explained above. Also in a similar manner
the curve of bending moment MM is obtained by finding
the area under the curve of shearing force. The maximum
ordinate of this curve will give the greatest bending moment
the ship will be subjected to under the assumed conditions.
In constructing curves of bending moment, moments
tending to cause " hogging " are put above the baseline, and
moments tending to cause " sagging " are put below the base
line. In the case in Fig. 99 the moments are " hogging "
throughout the whole length of the vessel.
The area of the curve of loads above the baseline being
the same as the area below the baseline, it follows that the
ordinate of the shearing force must come to zero at the end.
Also the ordinate of the curve of bending moment must be zero
at the end, and this constitutes a most effective check on the
accuracy of the work.
It is obvious, however, that the strains due to the bending
moment in still water are not the worst that in practice will
affect the longitudinal structure of the ship. The strains in still
water are small in magnitude compared with the strains that
2/0
Theoretical Naval Architecture.
may affect the ship at sea. For a ship at sea there are two
extreme cases that can be assumed, viz.
(1) The ship being supposed to be momentarily at rest on
the crest of a wave of her own length, the height of the wave
being taken some proportion of the length (Fig. 100).
(2) The ship being supposed to be momentarily at rest across
the trough of a wave similar to that assumed in (i) (Fig. 101).
It is usual to take the height of the wave from crest to crest,
or from trough to trough, ^ the length. The wave is assumed
to be of the form indicated by the " trochoidal theory," but no
account is taken of the internal structure of the wave. 1
Construction of a Trochoidal Wave Profile. The wave on
which a ship is supposed to be momentarily poised has for its
profile a curve called a " trochoid." This is a curve traced out
by a point inside a circle when the circle is rolled along a
straight line. The curve can be traced by its coordinates
referred to axes through the crest (Fig. QQA), viz.
T $ h a
x = L . . sin 9
27T 2
hi /l\
y = (i cos 0)
6 being given the values of 30, 60, etc., in circular measure.
L is the length of wave from crest to crest, and h is the height
from crest to trough. The curve may also be drawn by the
FIG.
construction indicated in Fig. gg\. It is noticed that the
curve is sharper at the crest than in the trough, which is a
characteristic feature of seawaves.
In order to get the displacement of the ship when on the
1 See the "Manual of Naval Architecture," by Sir W. H. White,
chaps, v. and viii. ; and a paper at the Institution of Naval Architects, by
Mr. (now Sir) W. E. Smith, M.I.N.A., in 1883; see also later.
Strains experienced by Ships, etc.
271
wave, it is convenient at each square station to run in a curve
of sectional areas. Then, if the profile of the wave is traced
and put on the profile of the ship, the area of each section up to
the surface of the wave is at once measured off, and these areas
integrated throughout the length give the displacement and
centre of buoyancy. If these are not correct, a further trial must
be made ; the ship may
have to be trimmed to
get the C.B. right, it
being essential that this
is in the same section
as the C.G.
With these assump
tions we can proceed
to construct the " curve
of buoyancy" for both
cases (i) and (2), and
from it and the u curve
of weights " we obtain
the "curve of loads."
Then, by the principles
explained above, we can
determine the " curve of
shearing force " and the " curve of bending moment." In Fig.
100 is given a set of curves for a ship on a wavecrest, and in
Fig. 1 01 a set of curves for the same ship astride a wavetrough.
In any case the maximum bending moment may be expressed
in the form wei g ht X len g th an( j j t j s foun d t h a t the value of
coefficient
this coefficient will not usually fall below 20 for either of the
extreme cases taken above. The maximum bending moment
in foottons for ordinary ships may be generally assumed at
from ^ to the product of the length in feet and the dis
placement in tons. The latter is frequently taken as a
standard value. In regard to this matter Mr. Foster King
made the following remarks at the I.N.A., 1915: "For
all ordinary vessels such as those with which we have to
deal, and many extraordinary ones, it has been found that
a close approximation to the greatest bending moment
FIG. 100.
272
Theoretical Naval Architecture.
obtained from direct calculations is obtained from the formula
, where L is the length, B the breadth, and
35 X 35
D is winter draught in feet.
FIG. 101.
Twenty years' experience of the
application of this for
mula has shown an
agreement of the order
of about 5 per cent, with
the calculated figures
furnished by builders
for vessels of the most
widely dissimilar cha
racter and dimensions
and give evidence of its
utility as a guide in
strength investigations."
This maximum bending
moment will usually
occur somewhere in the
vicinity of the midship
section.
It is usual to draw the distribution of weight as a series of
steps. For a large ship the length can be divided into sections
bounded by the main bulkheads, and the weights in each section
grouped together and plotted as uniform over each section.
Where extreme accuracy is desired, the work can be done in
greater detail. Fig. IOIA gives the curves for a torpedoboat
destroyer lying in the trough of a wave of height ~ her length,
all the bunkers being full. The use of a machine called the
" integraph " is of great value in getting quickly the curves of
shearing force and bending moment from the curves of loads.
If the pointer of this machine is run round the boundary of
a curvilinear area, the pen traces out a curve the ordinates of
which give the area of the original up to corresponding points.
This is just what is required in the above calculations. A
paper on the integraph was given by Mr. Johnson at the
Scottish Shipbuilders in 1904, to which the student is referred
for further information. See also a paper by Professor Biles
before the I.N.A. in 1905, for an exhaustive discussion on the
Strains experienced by Ships^ etc.
273
strength of ships, with special reference to calculations and
experiments on H.M.S. Wolf.
Stress on the Material composing the Section.
Considering now the ship's structure as a girder, a hogging
FIG. IOIA.
moment produces tension in the upper portion of the girder and
compression in the lower portion of the girder, the reverse being
true for a sagging moment.
We now have to consider in some detail how a given beam
T
2/4
Theoretical Naval Architecture.
is able to withstand the stresses on its material when subjected
to a given bending moment. Take a beam bent as in Fig. 92
and Fig. 102. AB is a longitudinal section and LL is a
transverse section of the beam. The upper layers are shortened
and the lower layers are lengthened. There must be one inter
mediate layer which is unaltered in length. This layer is
FIG.
called the " neutral surface," and the transverse section SS is
called the " neutral axis." This neutral axis can be shown to
pass through the centre of gravity of the section. 1 The bending
moment at the section LL is resisted by the compressive
stresses in the upper layers and the tensile stresses in the
lower layers.
It can be shown x that the following relation holds :
M
where / is the stress in tons per square inch at distance y
inches from the neutral axis.
M is the bending moment at this section in inchtons.
I is the moment of inertia of the section about the
neutral axis in inchunits.
It is by this formula that the stress on a particular portion
of the section of a beam can be determined, when we know the
1 See any standard work on " Applied Mechanics," as that by Professor
Cotterill, F.R.S.
Strains experienced by Ships, etc. . 275
bending moment at that section, the position of the neutral
axis, and the moment of inertia of the section about the neutral
axis.
Take, for example, the various sections of beams in Fig. 93.
Length of beam 12 feet, beam loaded in the middle with i ton
(neglecting the weight of the beam).
For (a) y = i"
1 = ^x16x4 = $ in inchunits
M = 36 inchtons
/. p = the stress at the top or bottom
36 X 12
= I X ^? 
64
= 6' 75 tons per square inch
For (*) y = 2"
I = ^ in inchunits
M = 36 inchtons
the stress at the top or bottom
3*375 tons per square inch
For (<:) y = 4"
I 2jjfl
in inchunits
M = 36 inchtons
/ = the stress at the top or bottom
1*6875 tons P er sc l uare
For (rf) y = 4
I = *
$* in inch units
M = 36 inchtons
/ = the stress at the top or bottom
= I '02 tons per square inch
Or looking at the question from another point of view, if
we say that the stress on the material is not to exceed 10 tons
per square inch, then we can determine for each of the sections
in Fig. 93 the greatest bending moment to which the beam
can be subjected.
For (a) M =  xl = ^xf$ = 53g inchtons
jr or () M=Xl = Tj Q X^j <5 = lo6l inchtons
276 Theoretical Naval Architecture.
For (c) M = ^ x I y X 2s = 2I 3 J inchtons
For (</) M=^xl = x*i* = 353$ inchtons
It is thus seen that the ratio of the bending moments that
these beams can stand is'
53 i :io6 : 2i 3 i :.353
or
1:2:4:6!.
The area of each section is the same, the only difference being
in the different distribution of the material of the section with
reference to the neutral axis.
We come now to the case of a ship subjected at a particular
section to either a " hogging " moment or a " sagging " moment.
To determine the stress on any portion of the section, we con
sider the vessel to be a large beam subjected to a given
bending moment, and we apply the formula
There are two things to be found before we can apply this
formula to a given section, viz. :
(i.) The position of the neutral axis, which passes through
the centre of gravity of the section.
(ii.) The moment of inertia of the section about the neutral
axis.
In considering the strength longitudinally of a section,
account must be taken only of such material as actually con
tributes to the strength through an appreciable length in the
vicinity of the section, such as plating of the inner and outer
bottom, keel, continuous longitudinals or keelsons, stringers,
deckplating, planking, etc.
A distinction must be made between material in tension and
material in compression. In tension, allowance must be made
for the material taken away from the plating, etc., for the rivet
holes, but in compression this deduction is unnecessary. It is
also usual to consider that wood is equivalent to y^ ^ ts area
in steel for both tension and compression. For an armoured
Strains experienced by Ships, etc.
277
vessel the armour is not assumed to take any tension, but is
assumed to be effective against compression.
We must accordingly have two separate calculations one
for the section under a hogging moment, and one for the
section under a sagging moment. The position of the neutral
axis and the moment of inertia of the section about the neutral
axis will be different for each case.
With reference to the above assumption, the following
remarks of Dr. Bruhn (I.N.A., 1899), wno ^ as given great
attention to this subject, may be noted :
Dr. Bruhn thinks that the correction for the rivetholes in the calcula
tion for I is more an act of error than of correction. This correction
assumes the structure highly discontinuous, the I and the position of the
neutral axis varying at the frame and between the frames. But the whole
theory of bending is based on the assumption that the structure is con
tinuous, and the bending certainly must be continuous. The I should
therefore be taken for the solid section^ and if it is desired to find the stress
between the rivets, we may increase the stress in the ratio in which the
sectional area is reduced (say g). This method for other than armoured
ships only requires one calculation for the moment of inertia, and when
dealing with shearing stresses, there is no more reason for deducting rivet
holes on one side than on the other.
The following form may conveniently be used for calculating
the position of the neutral axis and the moment of inertia of the
section about the neutral axis, areas being in square inches and
levers in feet :
I
Items.
2
Effective
area in
square
inches.
3
Lever
in feet.
4
Moment.
5
Lever
in feet.
6
Moment
of
Inertia.
7
A X a A x
h>
A
M
I
9
A section of the ship should be drawn out to scale with all
the scantlings shown on. An axis is assumed at about ctaehalf
the depth of the section. The several items are entered in
278  Theoretical Naval Architecture.
column i, the effective areas in column 2, and the distances of
the centres of gravity from the assumed axis in feet are entered
in column 3. For the items below the axis these levers are
negative. We thus obtain column 4, which gives the moment
of each item about the axis, and the algebraic sum of this
column, M, divided by the addition of column 2, viz. A, gives
the distance of the neutral axis from the assumed axis, in feet,
say d feet.
We now place in column 5 the same levers as in column 3,
and multiplying the moments in column 4 by the levers in
column 5, we obtain the areas of the several items multiplied
respectively by the square of their distances from the neutral
axis. Each of these products is, of course, positive. All these
added give a total I, say. For the portions of the section which
are vertical, an addition is needed for the moment of inertia of
the items about axes through their own centres of gravity, viz.
Y5 . A . fi 2 (see p. 104). For portions of the section which are
horizontal, h is small, and this addition may be neglected. We,
therefore, arrive at the moment of inertia of the section about
the assumed axis, viz. I + i" = I A , say. We now have to transfer
this moment of inertia about the assumed axis from that axis to
the neutral axis, or I = I A  A X d* y as explained on p. 104.
We can now determine the stress on the point of the section
farthest from the neutral axis, as this will be the point at which
the stress is greatest, by using the formula
M
The stresses thus found are only comparative, and must
be compared with those for a ship on service found to show
no signs of longitudinal weakness. Large ships can bear a
stress of 10 tons per square inch, because the standard wave is
exceptional. Ships 300 to 400 feet long have stresses 6 to 7
tons per square inch. In the special case of the Lusitania^
a stress of 10 tons per square inch was allowed when on a
wave of her own length of height onetwentieth the length.
It is to be observed that such a wave is of quite phenomena)
size, and unlikely to be encountered.
Strains experienced by Ships, etc.
279
Specimen Calculation for the Moment of Inertia,
etc., of Section. The following specimen calculation for a
torpedoboat destroyer is given as a guide to similar cal
culations. The depth of the girder was 17*7 feet and the
assumed neutral axis was taken as 9 feet above underside of
flat keel.
The column for areas was filled in with weight per foot run,
the total being turned into areas at the end (/ (area) means,
proportional to area, i.e. a function of area).
Above assumed neutral axis.
One side only.
W
/(Area)
Ibs. per ft.
(2)
Lever.
(3)
/(Moment).
(4)
/(I),
i.e. ( 2 )X( 3 )
(5)
d
(6)
Deck stringer .
67*5
828
559
4630
Remainder deck .
74*38
861
640
55 10
Girder coaming .
I40
8'95
1236
II2I
I'7S
43
Girder upper
angles . . .
80
9*24
740
68 3
Girder deck angles
30
884
265
234
Girder lower
angles . . .
7'
806
564
455
2nd girder plate .
2nd girder angles
4*2
80
817
817
} 996
813
3rd girder plate .
3rd girder angles
42
80
807
807
} 98*4
793
Bunker bulkhead
top plate . .
24*0
67
1607
1079
3*o
216
Bunker bulkhead
2nd plate . .
i6'8
3*92
65*9
2585
28
132
Bunker bulkhead
3rd plate . .
171
122
208
25*5
285
139
Gunwale angle .
7*o
800
560
448
Sheer strake . .
63*0
5'8o
365*3
2120
4*5
1275
Strake below . .
4095
158
64*7
102
4*55
847
Side stringer . .
897
3*03
272
82
3761 24381 18,354 12)2652
221 
iS.ot; 221
fl for tension 3079
1995 15,205
280 Theoretical Naval Architecture.
Below assumed neutral axis.
One side only.
AArea)
Ibs. per ft.
Lever.
/(Moment).
/(I)
d
y(A.</).
D strake .
3672
239
877
210
38
53 J
C strake .
46*0
555
255
I42O
29
387
B strake .
44'55
762
340
2590
i"4
87
A strake .
5562
858
477
4095
08
35
i Flat keel
300
9*0
270
2430
} Vertical keel
90
82
7T8
60 5
i '5
20
Lower angles
70
887
621
550
\ Upper angles
I Rider plate
60
833
752
742
6r8
339
457
ist long, angles
60
7 '4
44'4
328
2nd long, plates
536
69
37 'o
255
067
2
2nd long, angle
30
72
216
155
2nd long, angle
50
6.6
33"0
218
Lower side
stringer . .
897
i '37
123
17
Bilge keel plates
I0'0
66
660
435
08
6
Bilge keel angles
II'O
625
687
430
Bunker bulkhead
168
i '47
247
36
28
*32
Bunker bulkhead
32*0
473
I5i'5
717
40
5 12
34I35
21317
15,288
12)1711

143
I5.43I
'43
for tension 279 1743 12,630
Hogging. For hogging the full area below axis is taken,
and T 9 T the full area above axis to allow for rivet holes
/(Area) = 3079 + 34i'3 = 649*2
area = X 6492 = 3819 sq. in.
N.A. below assumed! _ 21317 1995
axis J 6492
/(I)aboutassumedaxis= 15,205 4 15,431 = 30,636
0*2 1 ft.
I about N.A. =   [30,636  (6492 x 02 1 2 )] = 18,000
3'4
N.A. above keel = 879 ft. N.A. below deck = 891 ft.
Bending moment = 10,275 ft. tons
Tensile stress in deck = ~  x 8*91 = 51 tons sq. in.
10,000
Compressive stressing 10,275
}   x 879 = 502 tons sq. in.
keel J 18,000
Sagging. For sagging the full area above axis is taken, and
the full area below axis to allow for rivet holes.
Strains experienced by Ships, etc. 281
/(Area) = 3761 + 279 = 6551
area = ^X 6551 = 385*4
J T"
N.A. above assumed! __ 2438  1743 __ ro6 ft
axis J 6551
/(I) about assumed axis = 18,575 + 12,630 = 31,205
I about N.A. = 4 [31,205  6 55 >]C X i'o6 2 ] = i7,9 20
o 4
N.A. above keel = 10*06 ft. N.A. below deck = 764 ft.
Bending moment = 13,000 ft. tons
Tensile stress in keel = ~ X 1006 = 73 tons sq. in.
Oppressive stress in) 13,000 fi = tons ^
deck J 17,920
Equivalent Girder. Although not necessary for calcu
lation purposes, it is frequently the practice to draw out for
HOGING. 
FIG. 103. FIG. 104.
the case under consideration a diagrammatic representation of
the disposition of the material forming the section. Such a
diagram will show at once how the material is disposed relative
to the neutral axis, and gives the section of the girder that
the ship is supposed to be. Such a diagram is termed the
" equivalent girder," and there must be one for hogging and
282
Theoretical Naval Architecture.
one for sagging, as shown in Fig. 103 and Fig. 104 respectively,
which are the equivalent girders for an armoured battleship.
A number of examples of equivalent girders for merchant
ships are given in Mr. Foster King's paper, I.N.A., 1913.
Shearing Stresses. We have seen above that to dispose
the material of a ship so as to resist most effectively the stresses
due to a bending moment, we must pay special attention to
the upper and lower portions of the girder. There are, how
ever, also shearing stresses in a loaded structure which, under
certain circumstances, may cause straining action to take place.
Professor Jenkins called attention to these stresses in a paper
before the I.N.A. in 1890, and their importance has been
increased in recent years owing to the great increase in the
size of the vessels built.
It is necessary first to deal with the shearing stresses which
occur in an ordinary loaded beam. In a beam, besides the
bending moment at each section, there is a tendency for each
section to slide over the adjacent one. This is measured by
the " shearing force." At each point of the section there is a
shearing stress set up to resist this sliding tendency. It can be
shown that such shearing stress is always accompanied by a
shearing stress of equal intensity on a plane at right angles.
Consider two consecutive sections of a beam K' and K" Sx apart
(Fig. IO4A), at which the^bending moments are M and M + 5M respec
K' K"
FIG. 1 04 A.
tively. Then, if/ is the normal stress at section K' at a distance y from
the neutral axis, and 5A is the element of area on which this stress acts,
the total normal force on the section above AB is
VI I
on proceeding to the limit. If now m is the moment of the area above AB
Strains experienced by Ships, etc. 283
about the neutral axis, the normal force is y m. At the consecutive
section this will be . m. The difference between these is the
resultant horizontal force on the portion of the beam 5* long above AB,
viz. . 5M. This must be the shearing force causing the section above
CD to slide along. If, now, q is the intensity of the shearing stress
along CD, and b the breadth of AB, we have q . b . 5x . . SM, or
in th j ^
ax L . b dx
the shearing force on whole section, so that we have q ' . This
shearing stress is therefore zero at the top and bottom of the section, and
it will vary at other points of the section.
To illustrate the variation of this shearing stress, take a beam of I
section or hollow section, as Fig. IO4A, subjected to a total shearing force
of 10 tons. It will be found that the variation of the shearing stress is
represented by the righthand portion of figure. It takes a sudden jump
at bottom of flange from 0*25 to 1*24 tons per square inch, because the
breadth is suddenly diminished from 5 inches to I inch. The maximum
value is 1*56 tons at the neutral axis. It is thus possible, in a beam with
a thin web, that an excessive shearing stress may be set up, and just at
that part of the section where there is no stress due to the bending moment.
It is also to be noticed that the shearing strength of steel is about four fifths
the tensile strength.
In a ship the shearing force amidships is usually zero (see
Figs. 100, 101), and the shearing force reaches a maximum at
about a quarter the length from each end. This, therefore,
will be the portion at which shearing strains are likely to be
most severe, and the maximum strains will occur in the neigh
bourhood of the neutral axis, because here the breadth of the
section is usually only twice the thickness of the bottom plating.
This stress has shown itself by the working of the rivets in
these portions of large ships, and the stresses vary also in
opposite directions according as the ship is in the trough or
on the crest of a wave. It is, therefore, becoming the practice
to work trebleriveted foreandaft laps at about middepth in
the fore and after bodies of large ships. 1
There are also setup stresses due to bending of the plating
owing to the varying pressure of water, and these, together with
1 Lloyd's Rules say: "In vessels of 480 feet and upwards, with side
plating less than 0*84 inch in thickness, the landing edges are to be treble
riveted for onefourth the vessel's length in the fore and after bodies for a
depth of onethird the depth."
284 Theoretical Naval Architecture.
the above stresses, have shown themselves by working in the
parts above mentioned, which has required the special strength
ening referred to. (The subject has been exhaustively discussed
by Dr. Bruhn in a paper before the Scottish Institution of Ship
builders, 1902.)
Principal Stress. When the material of a beam is
subjected to a tensile or compressive stress, together with a
shearing stress, these combine together to produce what is
termed the " principal stress " at any particular place. It can
be shown that if / is the ordinary tensile or compressive stress,
and q is the shearing stress, then the principal stress/ is given
by the equation
Thus, take a place immediately beneath the deck of a ship
with the stern overhanging in dry dock; A = 3' T > ^ = x '4 2 
Then/ = 3*65 tons, which is seen to be greater than the simple
tensile stress.
Unsymmetrical Bending. In the ordinary investiga
tion we assume that the ship is upright. If a ship is inclined
the depth of section is increased, and it may possibly happen
that an increased stress would be experienced at the corner of
the section.
Let MM (Fig. 1043) be the axis of the bending moment,
the ship being heeled to an angle 0. Then M, the bending
FIG. 1048.
moment, may be resolved into M . cos 6 with the axis OX, and
M . sin 6 with the axis OY, O being the C.G. of section. Each
Strains experienced by Skips, etc. 285
of these will produce stress at P as if it acted alone, and
the total stress at P will be
I 1} I 2 being the moments of inertia about axes OX, OY. The
position of the neutral axis is where / = o, or where  =  1
x 1 3
X tan 6 = tan <f>. If the point farthest from NN has coordinates
y and x' referred to OX, OY, then the maximum stress is
x' \
^ . cos 6 + = . sin 6 }
AI 2
Professor Biles (Scottish Shipbuilders, 18934) gives the
results for a ship for all angles from o to 90. He found, for
the ship he took, that the maximum stress was reached at 30,
and was there 20 per cent, greater than when the ship was
upright.
The " Smith " correction due to taking account
of the Internal Structure of a Wave. 1 In order to
understand this it will be necessary to deal with some features
of the trochoidal wave theory.
A trochoid is a curve traced out by a point inside a circle
when the circle is rolled along a straight line. Its coordinates
relative to axes through the crest are (Fig. 99A)
_ k u H /}!
x ~ 27r ~ 7 S1 1 I where L and H are the length
H I and height of the trochoid
y= (i cos0) }
If we imagine the circle rolled along with a velocity v, and
then a backward velocity v impressed on the system, we have
the points at extremities of tracing arms moving with uniform
angular velocity, and the wave formation will have a velocity
v. Fig. 1040 shows how the wave travels due to the revolu
tion anticlockwise of the points P. The particles in the crest
move in the same direction as the wave advance, and in the
trough in the opposite direction. A wave is the passage of
motion, and there is only a relatively small actual motion in
circular orbits of the particles of water composing the wave.
1 See a paper by Mr. (now Sir) W. E. Smith, I.N.A. 1883.
286
Theoretical Naval Architecture.
If R be the radius of the rolling circle then L = 27rR, and
= g . (dimensions in feet and seconds). The line of
FIG. 1040.
orbit centres is above the level of still water an amount J .
' R
where r is the radius OP for the surface trochoid or J . H (see
Fig. 1040).
For subsurfaces (see Fig. io4E) the rolling circle is the
FIG. 1040.
same. The axis is above the level of the same particles in
still water an amount i .  where r is the radius for the sub
j\.
surface in question.
For a subsurface the centre of whose rolling circle is at a
distance y below that of the surface trochoid, we have
r = r . e~&, where e is the base of Napierian logarithms =2718.
The values of r for values of y = i, 2, 3, etc. are therefore
_
(y = 2), r 2 = r . e R
(y = 3), r, = r Q .<rl' etc '
To evaluate these, we take logarithms first to the base
log. n = log. r  R , since log, e = i
log. r a = l og. r ^, and so on.
Strains experienced by Ships, etc.
28;
288 Theoretical Naval Architecture.
The pressure at any point in a trochoidal wave is the same
as at the point it occupies when in still water. Thus in Fig. io4E,
along the subsurface BB, the pressure is the same as at its still
water level b, and not due to its distance below the surface
trochoid (as is assumed in the standard method of calculation).
We therefore draw the trochoidal profile of the wave we
have to deal with and also the subsurfaces corresponding to
lines of orbit centres at distances say 2 feet apart. We then
calculate the positions of the corresponding stillwater levels,
a, b, c, d, <?,/.
Take as an illustrative example the wave drawn in Fig. io4E,
60 feet long, and 1 2 feet high. (This is, of course, an exagge
rated ratio of H 4 L, but has been selected for the sake of
clearness). The lines of orbit centres have been taken at
2 feet intervals, and the profile of the surface trochoid can be
drawn as already described In this case R = = 955,
and r = 6, being onehalf of the height of the surface trochoid.
In order to draw the subsurface trochoids we have to find the
radii for values of y = 2, 4, 6, etc.
We have log, r = log, r Q ^
Turning into ordinary logarithms to the base 10 (See Appendix
B.) we have
2*3 R
Putting in values of r and R, and successive values of y = 2,
4, 6, 8, 10, we have for values of r\ 486, 394, 3*19, 2'6 and
2' i. These will be the half heights of the subsurface trochoids
which can then be drawn in as indicated in the figure. The
level of still water below the lines of orbit centres is then
obtained by putting in the values of r in the expression ,
2K
or r88, 124, o'8i, 053, 035, 0*23, which are set down at
the side of the figure giving the levels a, b, c, d, <?, /. Thus
the pressure at any point on the subsurface DD is that due
to the distance of d below a. At the crest, if the wave pres
sures were neglected, the pressure at D would be that due to
Strains experienced by Ships, etc. 289
a head of 8*8 feet, whereas really it is only a pressure due to
a head of 47 feet.
Take now a ship on the wave and consider the section
that comes at HH. The levels of the surface A, and the sub
surfaces B, C, D, etc., are placed on the section (Fig. io4F),
A, B, C, etc. At the level of B we set up bb' = ab, at fc he
level of C, cc' = ac> etc., and a curve through ab'c', etc., will
give the line to work to to obtain the true value of the
buoyancy due to the section. In the case of sections in the
crest portion this is less than that up to the level of the wave.
Similarly for a section as at EE (Fig. 1040), where the value
of the effective buoyancy is greater. This done for sections
all along the length will result in a curve of effective areas.
When the ship is on the crest this curve is as dotted in
(Fig. io4H), and in the trough as dotted in (Fig. 104;), as
compared with the full curves obtained in the standard method.
These new curves of buoyancy must, of course, satisfy the
ordinary conditions, viz. that the displacement and position
of the centre of buoyancy are correct.
Trochoidal Wave Theory. The following are the
principal formulae arising out of the trochoidal wave theory,
where
L is length of wave in feet (crest to crest).
H is height of wave in feet.
T is periodic time in seconds, i.e. time of traversing the
length.
V is velocity of wave in knots (i.e. 6080 feet per hour).
v is velocity of wave in feet per second.
R is radius of rolling circle in feet.
; is radius of tracing arm in feet, i.e. onehalf the height
of wave = H.
r is radius of tracing arm in feet at depth of y where
y is depth of line of orbit centres from that line for the
surface trochoid.
g is acceleration due to gravity, viz. 322 in foot second
units.
V Q is a velocity of a particle in its orbit in feet per
second.
Theoretical Naval Architecture.
Section, EE
FIG. 1040.
FIG. io4H.
FIG. T<>4j.
Strains experienced by Ships^ etc. 291
r = ^ *=*. R = ^.L = 51 2L
V 2 =rSL, since V = z,x
r r d e~K, where e = 2718 log e e = i
r = Iog 10 r Q 'jp7
(since log, A = 23 Iog 10 . A).
log, r = log, r  j, and log, r = Iog 10 r Q 'jp7' g
2vrr TT.H H
v Q = y . v # for surface particle = y X # = 7*1 7^
Height of centre of orbits of a given particle above the level
of that particle in still water = ^ . For the wave surface
7T.H 2
this distance = ^ .
Rr
Virtual gravity at crest = 5 . .
R + r
at trough = . g.
EXAMPLES TO CHAPTER VII.
1. Determine the maximum stress on the section of an iron bar, 2
inches square and 20 feet long, when supported at the ends and unloaded
with one side horizontal. Ans. 6000 Ibs. per square inch.
2. An iron bar of the same length, and supported as in the previous
question, is of circular section, 2 inches diameter. Determine the maximum
stress. Ans. 8000 Ibs. per square inch.
3. A vessel floating in still water is subjected at a certain section to a
bending moment of 144 foottons. Determine the longitudinal stresses
(in pounds per square inch) in the material at top and bottom of this
section, assuming the section to be rectangular, 21 feet wide, 10 feet deep,
J" thick, and that the whole of it is effective in resisting stresses.
Ans. 223 Ibs.
4. The buoyancy of a vessel is o at the ends and increases uniformly to
the centre, while the weight is o at the centre and increases uniformly to
the ends. Draw the curves of shearing force and bending moment, and
find the maximum values of these quantities in terms of the displacement
and length of the vessel.
Ans. \ W, & W.L.
5. Suppose the skin and plate deck of an iron vessel to have the
following dimensions at the midship section, measured at the middle of
the thickness of the plates. Find the position of the neutral axis and
moment of resistance to bending. Breadth 48', and total depth 24', the
292 Theoretical Naval Architecture.
bilges being quadrants of 12' radius. Thickness of plate g" all round, and
coefficient of strength / = 4 tons.
Ans. Neutral axis 13" above centre of depth.
Moment of resistance to hogging, 32,500 foottons.
i> sagging, 39,000
(Examples 4 and 5 are from "Applied Mechanics," by Professor
Cotterill, F.R.S.)
6. A ship on a wavecrest is subjected at the midship section to a
hogging moment of 28,000 foottons. The depth of the section is 375
feet, and the neutral axis is 18*2 feet from the bottom, the moment of
inertia of the section about the neutral axis is 477,778 (square inches
X feet 2 ). Determine the maximum compressive and tensile stresses.
Ans. 1*07 tons per square inch compressive at bottom of section.
1*13 tons per square inch tensile at upper part of section.
7. State the maximum bending moment (in terms of weight and length
of vessel) in the case of a vessel having the weight uniformly distributed
and the curve of buoyancy a parabola. State also the position where
these maxima occur.
W
Ans. S.F. at o'2iL from end = .
105
B.M. amidships =
8. In question 3, if the ship has a bulwark each side 2 feet high, J inch
thick, what will then be the maximum stress ? Explain the significance
of your result as applying to actual ships. Ans. 263 Ibs.
Increase of depth of section will not necessarily diminish the maximum
stress.
.
y I p y I
and 8/ will be negative, i.e. stress diminishes only if > ^ .
This point acquires special importance 'in vessels with a light con
tinuous superstructure, as
(1) Boat deck in large cruisers.
(2) Superstructure in merchant vessels.
If the structure is made continuous, it is found that the influence of the
increased depth is greater than the increased I, and thus greater stresses
are likely to be experienced by the superstructure than it can bear. For
this reason, either
(1) A sliding joint is made, so that the superstructure contributes
nothing to the structural strength ; or, preferably
(2) The superstructure is made an integral part of the ship's structure.
See a paper by Mr. Montgomerie, I.N.A., 1915.
It does not follow that material added to a section will diminish the
maximum stress. We have
M 9y 51
Suppose a small area a is added at a distance /from neutral axis, then
this axis will shift :
JLiL
A+a
Strains experienced by Ships, etc. 293
The new I about old axis = 1 + a ./
The new I about new axis = I + a ./ 2  (A + a}( j^^ J
... increment of I or 51 = a ./> 
_ af
and I = A . #
S/ is positive if l  >  or 
/.*. the stress increases at distance y if the added material is placed less
&
than from the neutral axis.
/
In a rectangular beam, if material be added less than \ the depth
from middepth the stress is increased.
If a square inches of plating, placed at a distance of h feet above the
top of the girder, is such as to give the san,e stress as before (i.e. is the
same), show that
S.A.J
A(y + h)* + I
This formula was used by Mr. Montgomerie in his paper before
I.N.A., 1915, on "The Scantlings of Light Superstructures."
New I _ I
New_y y
9. A rectangular vessel is 30 feet broad and 20 feet deep, and has deck
plating \ inch thick, sides and bottom inch. At a certain section it is
subjected to a hogging moment of 20,000 tonsfeet, and a shearing force
of 300 tons. Calculate in tons per square inch
(1) Maximum tensile and compressive stresses.
(2) Maximum shearing stress.
(3) Principal stress immediately under deck.
Ans. (i) 71, 49 ; (2) 142 ; (3) 72.
10. In the previous example, what would be the stresses if the deck
and bottom were J inch thick, and the sides \ inch ?
Ans. (i) 5, 5; (2) 262; (3) 586.
11. An I beam 8 inches deep and I inch thick, with flanges 5 inches
wide, overhangs a distance of 5 feet, and a weight of 5 tons is placed at
the end. Determine at the point of leaving support (tons per square inch)
(1) Maximum tensile and compressive stresses.
(2) Maximum shearing stress.
(3) Principal stress immediately below the upper flange.
Ans. (i) 85, 85; (2) 078; (3) 646.
12. In a vessel of 10,000 tons displacement and 450 feet long, the
maximum bending moment is 3 \jW.L. The depth of midship section is
39 feet, and the neutral axis for hogging is at 0*49 the depth from keel.
294
Theoretical Naval Architecture.
The I for hogging about the neutral axis is 4000 in foot units. Calculate
the maximum tensile and compressive stresses.
Ans. 5'i8, 4*97 tons per square inch.
13. A vessel of 3000 tons displacement and 360 feet long has a maximum
hogging moment of ^ W.L. The draught is 14^ feet, and the freeboard
to stringer is 7$ feet. The neutral axis for hogging is 3*15 feet below the
waterline, and the moment of inertia about neutral axis is 73,000 (square
inches X feet 2 ). What are the maximum tensile and compressive stresses ?
Ans. 5'i8, 5*41 tons per square inch.
14. A vessel is designed with a very large overhang at the stern from
the cutup of the keel. Indicate what calculations you would make to see
if the ship could safely be dry docked with the stern unsupported. Indi
cate how you would strengthen such a ship to withstand the strains set up
in dry dock.
15. Two similar vessels are respectively 300 feet long, 2135 tons dis
placement, and 360 feet long, 3000 tons displacement, the depth being
nearly the same. Indicate the method you would adopt to ensure the
second ship being strong enough, and estimate the increase of maximum
bending moment the second ship has to stand as compared with the first.
Ans. About 68 per cent, greater B.M.
1 6. The effective part of the transverse section of a vessel amidships is
represented by the diagram, the vessel being 42 feet broad and 28 feet deep.
Find the maximum tensile and
; compressive stresses when the vessel
is subjected to a sagging moment of
60,000 foottons. The plating is
\ inch thick and no allowance need
be made for rivet holes and laps of
plating. (Honours B. of E., 1908.)
This example is worked out
below. The best way to proceed is
to prepare a table similar to that in
this chapter ; attention is necessary to
the units, the areas being in square
inches and the lengths in feet. We
therefore have, taking in the first
place all distances from the keel
t
Items.
Area in
sq. in.
Levers
in ft.
Moment.
Levers
in ft.
Moment of
inertia.
A.A.*
Upper deck
Main deck .
252
252
28
2O
7,056
5>40
28
20
197,578
100,800
Tank top .
252
4
1, 008
4
4,032
Bottom
252
Sides . . .
Girders . .
336
48
M
2
4,704
9 6
14
2
65,856
I 9 2
21,952
6 4
1392
17,904
368,458
22,016
Strains experienced by Ships, etc. 295
368,458
17904 _ I2 . 86 ft c G from keel> 22,016
*39 2 390,474 I about keel in sq. in x ft.*
I about neutral axis = 390, 474  1392 x (i2'86) 2 = 160,224
stress at keel tensile = X 1 2 '86 4 '8 1 tons sq. in,
stress at deck compressive = l6o x I S' l 4 = 5' 6 7 ton s sq. in.
17. An Atlantic ocean wave is 600 feet long and 40 feet high. Cal
culate the radii of the orbits at depths of loo, 200, 300, 400, 500 and
600 feet.
Ans. 703, 249, 0*87, 0*31, o'li, 0*04 feet.
These results show that even in a wave of large dimensions at a depth
less than the length of the wave the motion of the water is practically nil.
1 8. The successive crests of the wave profile along a ship's side going
at speed in still water are observed to be about 300 feet apart. What is
the speed of the vessel in knots ?
Ans. 23 knots about.
19. What is the speed of a 600 feet wave in knots ?
Ans. 33 knots nearly.
20. What is the length of a wave successive crests of which are observed
to pass a stationary observer at intervals of 8 seconds ?
Ans. 330 feet.
21. A wave is 600 feet long and 40 feet high. Compare the orbital
velocity of the particles in the surface with the speed of the wave.
Ans. 1 1 '6 : 55 '4, or about \.
22. What is the virtual force of gravity in the crest and trough respec
tively of a wave 600 feet long and 40 feet high ?
Ans. 0.792, I'l ig.
CHAPTER VIII.
HORSEPOWER, EFFECTIVE AND INDICATED RESIST
ANCE OP SHIPS COEFFICIENTS OF SPEED LA W OF
COMPARISON PROPULSION.
Horsepower. We have in Chapter V. defined the " work "
done by a force as being the product of the force and the
distance through which the force acts. Into the conception
of work the question of time does not enter at all, whereas
" power " involves not only work, but also the time in which
the work is done. The unit of power is a "horsepower?
which is taken as " 33,000 footlbs. of work performed in i
minute? or "550 footlbs. of work performed in i second?
Thus, if during i minute a force of i Ib. acts through 33,000
feet, the same power will be exerted as if a force of 33 Ibs. acts
through 1000 feet during i minute, or if 50 Ibs. acts through
ii feet during i second. Each of these will be equivalent to
i horsepower. The power of a locomotive is a familiar in
stance. In this case the work performed by the locomotive
if the train is moving at a uniform speed is employed in
overcoming the various resistances, such as the friction of the
wheels on the track, the resistance of the air, etc. If we
know the amount of this resistance, and also the speed of the
train, we can determine the horsepower exerted by the loco
motive. The following example will illustrate this point :
If the mass of a train is 150 tons, and the resistance to its motion
arising from the air, friction, etc., amount to 16 Ibs. weight per ton when
the train is going at the rate of 60 miles per hour on a level plain, find the
horsepower of the engine which can just keep it going at that rate.
Resistance to onward motion = 150 X 16
= 2400 Ibs.
Speed in feet per minute = 5280
Work done per minute = 2400 X 5280 footlbs.
2400 x 5280
Horsepower =
33000
= 384
Horsepower, Effective and Indicated, etc. 297
In any general case, if
R = resistance to motion in pounds ;
v = velocity in feet per minute ;
V = velocity in knots (a velocity of i knot is 6080 feet
per hour) ;
then
1? ^^ <7J
Horsepower =
33000
X V) nearly
The case of the propulsion of a vessel by her own engines
is much more complicated than the question considered above
of a train being drawn along a level plain by a locomotive.
We must first take the case of a vessel being towed through
the water by another vessel. Here we have the resistances
offered by the water to the towed vessel overcome by the strain
in the towrope. In some experiments on H.M.S. Greyhound
by the late Mr. Froude, which will be described later, the tow
rope strain was actually measured, the speed being recorded
at the same time. Knowing these, the horsepower necessary
to overcome the resistance can be at once determined. For
example
At a speed of 1017 feet per minute, the towrope strain was 10,770 Ibg.
Find the horsepower necessary to overcome the resistance.
Work done per minute = 10,770 X 1017 footlbs.
TT 10770 X 1017
Horsepower =
33000
= 332
Effective Horsepower. The effective horsepower of
a vessel at a given speed is the horsepower required to over
come the various resistances to the vessel's progress at that
speed. It may be described as the horsepower usefully
employed, and is sometimes termed the " towrope " or " tug "
horsepower, because this is the power that would have to be
transmitted through the towrope if the vessel were towed
through the water at the given speed. Effective horsepower
is often written E.H.P. We shall see later that the E.H.P. is
entirely different to the Indicated Horsepower (written I.H.P.),
298 Theoretical Naval Architecture.
which is the horsepower actually measured at the vessel's
engines.
Example. Find the horsepower which must be transmitted through
a towrope in order to tow a vessel at the rate of 16 knots, the resistance to
the ship's motion at that speed being equal to a weight of 50 tons.
Ans. 5503 H.P.
Experiments with H.M.S. "Greyhound," by the
late Mr. William Fronde, P.R.S. These experiments
took place at Portsmouth as long ago as 1871, and they settled
a number of points in connection with the resistance and pro
pulsion of ships, about which, up to that time, little was known.
The thoroughness with which the experiments were carried
out, and the complete analysis of the results that was given,
make them very valuable ; and students of the subject would
do well to consult the original paper in the Transactions of the
Institution of Naval Architects for 1874. A summary of the
experiments, including a comparison with Rankine's "Aug
mented Surface Theory of Resistance," will be found in vol. ill
of Naval Science. Mr. Froude's report to the Admiralty was
published in Engineering, May i, 1874.
The Greyhound was a ship 172' 6" in length between per
pendiculars, and 33' 2" extreme breadth, the deepest draught
during the experiments being 1 3' 9" mean. The displacement
FIG. 105.
corresponding to this mean draught being 1161 tons; area of
midship section, 339 square feet ; area of immersed surface,
7540 square feet. The Greyhound was towed by H.M.S.
Active. It was essential to the accuracy of the experiments
that the Greyhound should proceed through undisturbed water,
and to avoid using an exceedingly long towrope a boom was
rigged out from the side of the Active to take the towrope (see
Fig. 105). By this means the Greyhound proceeded through
Horsepower, Effective and Indicated, etc. 299
water that had not been influenced by the wake of the Active.
The length of the boom on the Active was 45 feet, and the length
of the towrope was such that the Greyhound's bow was 190
feet clear of the Aciive's stern. The actual force on the tow
rope at its extremity was not required, but the " horizontal
component." This would be the force that was overcoming
the resistance, the "vertical component" being due to the
weight of the towrope. The horizontal force on the towrope
and the speed were automatically recorded on a sheet of paper
carried on a revolving cylinder. For details of the methods
employed and the apparatus used, the student is referred to
FIG. 106.
the sources mentioned above. The horizontal force on the
towrope was equal to the nett resistance of the Greyhound,
The results can be represented graphically by a curve, abscissae
representing speed, and ordinates representing the resistance
in pounds. Such a curve is given by A in Fig. 106.
It will be seen that the resistance increases much more
rapidly at the higher than at the lower speeds; thus, on
increasing the speed from 7 to 8 knots, an extra resistance
of 1500 Ibs. has to be overcome, while to increase the speed
3OO Theoretical Naval Architecture.
from ii to 12 knots, an extra resistance of 6000 Ibs. must
be overcome. Beyond 12 knots the shape of the curve
indicates that the resistance increases very rapidly indeed.
Now, the ratt at which the resistance increases as the speed
increases is a very important matter. (We are only concerned
now with the total resistance.) Up to 8 knots it was found
that the resistance was proportional to the square of the speed ;
that is to say, if R 1} R, represent the resistances at speeds
Vj, V 2 respectively, then, if the resistance is proportional to
the square of the speed
R, : R 2 : : V, 2 : V a '
By measuring ordinates of the curve in Fig. 106, say at 5 and 6
knots, this will be found to be very nearly the case. As the
speed increases above 8 knots, the resistance increases much
more rapidly than would be given by the above ; and between
1 1 and 1 2 knots, the resistance is very nearly proportional to
the fourth power of the speed.
The experiments were also conducted at two displacements
less than 1161 tons, viz. at 1050 tons and 938 tons. It was
found that differences in resistance, due to differences of
immersion, depended, not on changes of area of midship
section or on changes of displacement, but rather on changes
in the area of wetted surface. Thus for a reduction of 19^
per cent, in the displacement, corresponding to a reduction of
area of midship section of 16^ per cent., and area of immersed
surface of 8 per cent., the reduction in resistance was about
10^ per cent., this being for speeds between 8 and 12 knots.
Ratio between Effective Horsepower and Indi
cated Horsepower. We have already seen that, the
resistance of the Greyhound at certain speeds being deter
mined, it is possible to determine at once the E.H.P. at
those speeds. Now, the horsepower actually developed by the
Greyhound's own engines, or the " indicated horsepower "
(I.H.P.), when proceeding on the measured mile, was observed
on a separate series of trials, and tabulated. The ratio of the
Horsepower, Effective and Indicated, etc. 301
E.H.P. to the I.H.P. was then calculated for different speeds,
and it was found that E.H.P. 4 I.H.P. in the best case was
only 0*42 ; that is to say, as much as 58 per cent, of the power
was employed in doing work other than overcoming the actual
resistance of the ship. This was a very important result, and
led Mr. Froude to make further investigations in order to
determine the cause of this waste of power, and to see whether
it was possible to lessen it.
Tf TT T>
The ratio y Vj ' ' at any given speed is termed the "pro
pulsive coefficient" at that speed. As we saw above, in the
most efficient case, in the trials of the " Greyhound" this co
efficient was 42 per cent. For modern vessels with fine lines a
propulsive coefficient of 50 per cent, may be expected, if the
engines are working efficiently and the propeller is suitable.
In special cases, with extremely fine forms and fastrunning
engines, the coefficient rises higher than this. These values only
hold good for the maximum speed for which the vessel is
designed ; for lower speeds the coefficient becomes smaller.
The following table gives some results as given by Mr. Froude.
The Mutine was a sistership to the Greyhound^ and she had
also been run upon the measured mile at the same draught and
trim as the Greyhound.
1
JH
g
i .

1
T3 O hn
8<s sa
i
1
c
1'* B$
i
h .
Ship.
S
13 3
jl
?!:!
as u;
Sfll
o,2 crj
Ou t) fO
a
T3
II
ti
i
JM!O
S x
U
i) *^ *S U
d
ji
o
Illl
i
.
jj
1.
w
J
H
Greyhound
/ioi7
i 845
10,770
6,200
I587
453
0350
Mutine
/ 977
\ 757
9,440
4,770
2795
1094
770
328
0363
0334
3 02
Theoretical Naval Architecture.
Resistance. We now have to inquire into the various
resistances which go to make up the total resistance which a
ship experiences in being towed through the water. These
resistances are of three kinds
1. Resistance due to friction of the water upon the surface
of the ship.
2. Resistance due to the formation of eddies.
3. Resistance due to the formation of waves.
1. " Frictional resistance? or the resistance due to the
friction of the water upon the surface of the ship. This is
similar to the resistance offered to the motion of a train on a
level line owing to the friction of the rails, although it follows
different laws. It is evident that this resistance must depend
largely upon the state of the bottom. A vessel, on becoming
foul, loses speed very considerably, owing to the greatly
increased resistance caused. This frictional resistance forms
a large proportion of the total at low speeds, and forms a
good proportion at higher speeds.
2. Resistance due to eddymaking. Take a block of wood,
and imagine it placed a good distance below the surface of
a current of water moving at a uniform speed V. Then
the particles of water will run as approximately indicated
in Fig. 107 At A we shall have a mass of water in a state of
FIG. 107.
violent Agitation, and a much larger mass of water at the rear
of the block. Such masses of confused water are termed
"eddies" and sometimes "dead water." If now we imagine
that the water is at rest, and the block of wood is being towed
Horsepower, Effective and Indicated, etc. 303
through the water at a uniform speed V, the same eddies will
be produced, and the eddying water causes a very considerable
resistance to the onward motion. Abrupt terminations which
are likely to cause such eddies should always be avoided in
vessels where practicable, in order to keep the resistance as
low as possible. This kind of resistance forms a very small
proportion of the total in wellformed vessels, but in the older
vessels with full forms aft and thick sternposts, it amounted to
a very considerable item.
3. Resistance due to the formation of waves. For low
speeds this form of resistance is not experienced to any
sensible extent, but for every ship there is a certain speed
above which the resistance increases more rapidly than would
be the case if surface friction and eddymaking alone caused
the resistance. This extra resistance is caused by the forma
tion of waves upon the surface of the water.
We must now deal with these three forms of resistance in
detail, and indicate as far as possible the laws which govern
them.
i . Frictional Resistance. The data we have to work upon
when considering this form of resistance were obtained by the
late Mr. Froude. He conducted an extensive series of experi
ments on boards of different lengths and various conditions
of surface towed edgewise through water contained in a tank,
the speed and resistance being simultaneously recorded. The
following table represents the resistances in pounds per square
foot due to various lengths of surface of various qualities when
moving at a uniform speed of 600 feet per minute, or very nearly
6 knots in fresh water. There is also given the powers of the
speed to which the resistances are approximately proportional.
We can sum up the results of these experiments as follows :
The resistance due to the friction of the water upon the surface
depends upon
(1) The area of the surface.
(2) The nature of the surface.
(3) The length of the surface.
(4) The density of the water.
and (5) The resistance varies as the ** power of the speed
where n varies from 1*83 to 2*16.
304
Theoretical Naval Architecture.
L.HNGT
1 OF Su
UFACE
N FEET
a
8
S
to
JO
Si
it
si
i
1
I
43
K,
Nature of surfa ce.
^'5
"S .
;:
1
!*'3
1
"*'S
S
o o
3 "
O O
is
"Sti.
SJ
T3
y
~c
S.J
.a
US
1.5
2 S
s,
B U
8.S*
B V
. G
I 1
b C
It
II
*8
It
El
O'S
* S
1
1
2
11
I
w rt
If
.2
H
Powe
resista
I
Varnish ...
Tinfoil ...
200
216
041
030
I8 5
199
0325
0278
IS
0278
0*262
183
O25O
0246
Calico ...
*'93
087
192
0*626
1*89
053I
I8 7
0474
Fine sand
2"OO
081
200
0583
200
0*480
206
0405
Medium sand .
200
090
2'00
0625
200
0534
2*00
0488
And thus we can write for a smooth surface in salt water
where R = resistance in pounds ;
S = area of surface in square feet ;
V = speed in knots relative to still water ;
/ = a coefficient depending upon the nature and length
of the surface ;
w = density of salt water ;
w = density of fresh water ;
w/w = 1*025.
This coefficient /will be the resistance per square foot given in
the above table, as is at once seen by making S = i square foot,
V = 6 knots, and w = w . It is very noticeable how the resistance
per square foot decreases as the length increases. Mr. Froude
explained this by poirting out that the leading portion of the
plane must communicate an onward motion to the water which
rubs against it, and " consequently the portion of the surface
which succeeds the first will be rubbing, not against stationary
water, but against water partially moving in its own direction,
and cannot therefore experience as much resistance from it."
H or SBpower > Effective and Indicated, etc. 305
Experiments were not made on boards over 50 feet in
length. Mr. Froude remarked, in his " report, "It is highly
desirable to extend these experiments, and the law they eluci
date, to greater lengths of surface than 50 feet; but this is the
greatest length which the experimenttank and its apparatus
admit, and I shall endeavour to organize some arrangement by
which greater lengths may be successfully tried in open water."
Mr. Froude was never able to complete these experiments
as he anticipated. It has long been felt that experiments with
longer boards would be very valuable, so that the results could
be applied to the case of actual ships. It is probable that in
the new American experiment tank, 1 which is of much greater
length than any others at present constructed, experiments with
planes some hundreds of feet in length may be carried out.
These experiments show very clearly how important the
condition of the surface is as affecting resistance. The
varnished surface may be taken as typical of a surface coated
with smooth paint, or the surface of a ship sheathed with
bright copper, the medium sand surface being typical of the
surface of a vessel sheathed with copper which has become
foul. If the surface has become fouled with large barnacles,
the resistance must rise very high.
In applying the results of these experiments to the case of
actual ships, it is usual to estimate the area of wetted surface,
and to take the length of the ship in the direction of motion to
determine what the coefficient/ shall be. See below for E.H.P.
due to friction and eddymaking.
Take the following as an example :
The wetted surface of a vessel is estimated at 7540 square feet, the
length being 172 feet. Find the resistance due to surface friction at a
speed of 12 knots in salt water, assuming a coefficient of 0*25, and that
the resistance varies (a) as the square of the speed, and (b) as the 183
power of the speed.
(a) Resistance = 0*25 X I '025 X 7540 X () 2
= 7728 Ibs.
(b) Resistance = 0*25 X 1*025 x 7540 X (Jf) 1 ' 8 *
= 6870 Ibs. 2
1 For a description of this tank, see Engineering, Dec. 14, 1900. See
discussion on paper by Mr. Baker on Frictional Resistance, I.N.A., 1916.
2 T.his.hasto be obtained by the aid of logarithms.
306 Theoretical Naval Architecture.
It is worth remembering that for a smooth painted surface
the frictional resistance per square foot of surface is about \ Ib.
at a speed of 6 knots.
It is useful, in estimating the wetted surface for use in the
above formula, to have some method of readily approximating
to its value. Several methods of doing this have been already
given in Chapter II., the one known as " Kirk's Analysis "
having been largely employed. There are also several approxi
mate formulae which are reproduced
(1) Based on Kirk's analysis
Surface = 2 LD + Y
(2) Given by Mr. Denny
Y"
Surface = ryLD +
(3) Given by Mr. Taylor
Surface = 1 5 '5 VW.L.
(4) Used at the Experiment Tank at Haslar
Surface = '(3 4 f ,'
L being the length of the ship in feet ;
D being the mean moulded draught ;
V being the displacement in cubic feet ;
W being the displacement in tons.
2. Eddymaking Resistance. We have already seen the
general character of this form of resistance. It may be
assumed to vary as the square of the speed, but it will vary
in amount according to the shape of the ship and the appen
dages. Thus a ship with a full stern and thick sternposts
will experience this form of resistance to a much greater
extent than a vessel with a fine stern and with sternpost and
rudder of moderate thickness. Eddymaking resistance can
be allowed for by putting on a percentage to the frictional
resistance. It is possible to reduce eddymaking to a
minimum by paying careful attention to the appendages and
endings of a vessel, especially at the stern. Thus shaft
brackets in twinscrew ships are often made of pearshaped
Horsepower, Effective and Indicated, etc. 307
section, as shown in Fig. 85 E J and Fig. 108. A conical piece is
always put at the after end of propeller shafts for this reason.
AFT. FOR*
FIG. 108.
The following formula can be used to express the effective
horsepower due to surface friction and eddymaking in salt
water, viz. :
E.H.P. = ^./.S.V 2 ' 88
V being in knots.
For the coefficient/, we can take/ = 0009 f r a length of
500 feet varying to 0*01 for a length of 40 feet. These values
are rather greater than would be inferred from Froude's
experiments, and include an allowance for eddymaking
resistance.
On page 332, a table is given for the E.H.P. due to skin
friction, based on Mr. Froude's constants, assuming the skin
friction to vary as V 1825 , from speeds of 10 to 40 knots, and
for lengths of 100 to 1000 feet. The reduction of the co
efficient as length increases has been allowed for in this
table.
Mr. Baker, in his work on t " Resistance and Propulsion,"
gives the following values of f for salt water and for varying
lengths in the formula
Frictional resistance in Ibs. =/. S . V 1825
where S is wetted surface in square feet
V is speed in knots.
Length)
in feet /
50
75
100
200
300
400
500
700
900
/
00096
000935
00092
000898
00089
000883
000877
000868
00086
See note in " Strength of Shaft Brackets," p. 255, as to resistance of
308
Theoretical Naval Architecture.
These values are obtained by assuming that the frictional
coefficient of the first 50 feet is the same as that of a 5ofoot
plank, regardless of the ship's length, and that the remainder
of the length has the same frictional resistance as the last foot
of the 5ofoot plank.
The table given for E.H.P. due to frictional resistance per
square foot of wetted surface given on page 332 is calculated
from similar figures to the above, and it is suggested as
an exercise that some of the figures given be checked.
Attention is necessary to the units, as the above is for
resistance in Ibs., and E H.P. = ^ . R . V, so that
E.H.P. = ^./.V 2 ' 826
3. Resistance due to the Formation of Waves. A completely
submerged body moving at any given speed will only experi
ence resistance due to surface friction and eddymaking provided
FIG. 109.
it is immersed sufficiently ; but with a body moving at the
surface, such as we have to deal with, the resistance due to
the formation of waves becomes very important, especially at
high speeds. This subject is of considerable difficulty, and
it is not possible to give in this work more than a general
outline of the principles involved.
Horsepower , Effective and Indicated, etc. 309
Consider a body shaped as in Fig. 109 placed a long way
below the surface in water (regarded as frictionless), and
suppose the water is made to move past the body with a uniform
speed V. The particles of water must move past the body in
certain lines, which are termed streamlines. These stream
lines are straight and parallel before they reach the body, but
owing to the obstruction caused, the particles of water are
locally diverted, and follow curved paths instead of straight
ones. The straight paths are again resumed some distance at
the rear of the body. We can imagine these streamlines
making up the boundaries of a series of streamtubes, in each
of which the same particles of water will flow throughout the
operation. Now, as these streams approach the body they
broaden, and consequently the particles of water slacken in
speed. Abreast the body the streams are constricted in area,
and there is a consequent increase in speed ; and at the rear of
the body the streams again broaden, with a slackening in speed.
Now, in water flowing in the way described, any increase in
speed is accompanied by a decrease in pressure, and conversely
any decrease in speed is accompanied by an increase in pressure.
We may therefore say
(1) There is a broadening of all the streams, and attendant
decrease of speed and consequent excess of pressure, near both
ends of the body ; and
(2) There is a narrowing of the streams, with attendant
excess of speed and consequent decrease of pressure, along the
middle of the body.
This relation between the velocity and pressure is seen in
the draught of a fire under a chimney when there is a strong wind
blowing. The excess of the speed of the wind is accompanied
by a decrease of pressure at the top of the chimney. It
should be noticed that the variations of velocity and pressure
must necessarily become less as we go further away from the
side of the body. A long way off the streamlines would be
parallel. The body situated as shown, with the frictionless
\\ater moving past it, does not experience any resultant force
tending to move it in the direction of motion. 1
1 This principle can be demonstrated by the use of advanced mathematics.
" We may say it is quite evident if the body is symmetrical, that is to say,
3io Theoretical Naval Architecture.
Now we have to pass from this hypothetical case to the case
of a vessel on the surface of the water. In this case the water
surface is free, and the excess of pressure at the bow and stern
shows itself by an elevation of the water at the bow and stern,
and the decrease of pressure along the sides shows itself by a
depression of the water along the sides. This system is shown
by the dotted profile of the water surface in Fig. no, which
FIG. no.
has been termed the statical wave. The foregoing gives us
the reason for the wavecrest at the stern of the ship. The
crest at the bow appears quite a reasonable thing to expect,
but the crest at the stern is due to the same set of causes.
This disturbance of level at the bow and stern is described by
Mr. R. E. Froude as the " forcive " of the actual wave forma
tion. If a stone is thrown into water, the sudden disturbance
propagates a series of waves that radiate in all directions. In
the case of a ship, the shape of the ship causes the disturbance
to form diverging and transverse waves as seen below.
Observation shows that there are two separate and distinct
series of waves caused by the motion of a ship through the
water : (i) at the bow, and (2) at the stern.
Each of these series of waves consists of (i) a series of
diverging waves, the crests of which slope aft, and (2) a series
of transverse waves, whose crests are nearly perpendicular to
the middle line of the ship.
First, as to the diverging waves at the bow. " The inevi
tably widening form of the ship at her entrance throws off on
each side a local oblique wave of greater or less size according
to the speed and obtuseness of the wedge, and these waves
form themselves into a series of diverging crests. These waves
has both ends alike, for in that case all the fluid action about the after
body must be the precise counterpart of that about the fore body j all the
streamlines, directions, speed of flow, and pressures at every point must be
symmetrical, as is the body itself, and all the forces must be equal and
opposite" (see a paper by Mr. R. E. Froude, on "Ship Resistance," read
before the Greenock Philosophical Society in 1894).
Horsepower, Effective and Indicated, etc. 3 1 1
have peculiar properties. They retain their identical size for a
very great distance, with but little reduction in magnitude.
But the main point is, that they become at once disassociated
with the vessel, and after becoming fully formed at the bow,
they pass clear away into the distant water, and produce no
further effect on the vessel's resistance." These oblique waves
are not long in the line of the crest BZ, Fig. in, and the
waves travel perpendicular to the crestline with a speed of
V cos 0, where V is the speed of the ship. As the speed of
the ship increases the diverging waves become larger, and
consequently represent a greater amount of resistance.
Besides these diverging waves, however, " there is produced
by the motion of the vessel another notable series of waves,
which carry their crests transversely to her line of motion." It
is this transverse series of waves that becomes of the greatest
importance in producing resistance as the speed is pushed
to values which are high for the ship. These transverse waves
show themselves along the sides of the ship by the crests and
troughs, as indicated roughly in Fig. no. The lengths of these
waves (i.e. the distance from one crest to the other) bears a
definite relation to the speed of the ship. This relation is that
the length of the wave varies as the square of the speed at
which the ship is travelling, and thus as the speed of the ship
increases the length from crest to crest of the accompanying
series of transverse waves increases very rapidly.
The waves produced by the stern of the ship are not of
such great importance as those formed by the bow, which we
have been considering. They are, however, similar in character,
there being an oblique series and a transverse series.
3 I2
Theoretical Naval Architecture.
Interference between the Bow and Stern Transverse Series of
Waves. In a paper read by the late Mr. Froude at the Insti
tution Of Naval Architects in 1877, some very important
experiments were described, showing how the residuary resist
ance 1 varied in a ship which always had the same fore and
after bodies, but had varying lengths of parallel middle body
inserted, thus varying the total length. A strange variation in
the resistance at the same speed, due to the varying lengths of
parallel middle body was observed. The results were set out
as roughly shown in Fig. 112, the resistance being set up on a
340.
240 iW 40
LENGTH OF PARALLEL MIDDLE BODY
FIG. 112.
base of length of ship for certain constant speeds. At the low
speed of 9 knots very little variation was found, and this was
taken to show that at this speed the residuary resistance was
caused by the diverging waves only.
The curves show the following characteristics :
(1) The spacing or length of undulation appears uniform
throughout each curve, and this is explained by the fact that
waves of a given speed have always the same length.
(2) The spacing is more open in the curves of higher speed,
the length apparently varying as the square of the speed. This
is so because the length of the waves are proportionate to the
square of the speed.
1 Residuary resistance is the resistance other than frictional.
Horsepower, Effective and Indicated, etc. 313
(3) The amplitude or heights of the undulations are
greater in the curves of higher speeds, and this is so, because
the waves made by the ship are larger for higher speeds.
(4) The amplitude in each curve diminishes as the length of
parallel middle body increases, because the wave system, by
diffusing transversely, loses its height.
These variations in residuary resistance for varying lengths
are attributed to the interference of the bow and stern trans
verse series of waves. When the crests of the bowwave series
coincide with the crests of the sternwave series, the residuary
resistance is at a maximum. When the crests of the bowwave
series coincide with the troughs of the sternwave series, the
residuary resistance is at a minimum.
The following formula l gives an approximation to the effec
tive horsepower to overcome wavemaking resistance, viz.
The coefficient b, however, has varying values for the same
ship owing to the interference above mentioned, so that it is
not a formula that can be relied upon. The total formula for
E.H.P. can be written
E.H.P. = /. S . V 2 ' 83 + b . . V B
where / is a coefficient for surface friction and eddymaking
appropriate to the length. If 50 per cent, be taken as a
standard propulsive coefficient at top speed, to 40 per cent.
at 10 knots, say, values of b can be determined from trial data
in the user's possession which may be useful for estimating
purposes. Examples 31 and 32 in Appendix illustrate its use.
The following extracts from a lecture 2 by Lord Kelvin (Sir
William Thomson) are of interest as giving the relative in
fluence of frictional and wavemaking resistance :
" For a ship A, 300 feet long, 31^ feet beam, and 2634 tons
displacement, a ship of the ocean mailsteamer type, going at
13 knots, the skin resistance is 5*8 tons, and the wave resistance
1 See Mr. Johns' paper, I.N.A., 1907, for a discussion of "approxi
mate formulae for determining the resistance of ships," also Prof. Hovgaard,
1908, I.N.A.
9 Third volume " Popular Lectures and Addresses," 1887.
314 Theoretical Naval Architecture.
is 3*2 tons, making a total of 9 tons. At 14 knots the skin
resistance is but little increased, viz. 6*6 tons, while the wave
resistance is 6*15 tons.
" For a vessel B, 300 feet long, 463 feet beam, and
3626 tons, no parallel middle body, with fine lines swelling out
gradually, the wave resistance is much more favourable. At
13 knots the skin resistance is rather more than A, being
6*95 tons as against 5*8 tons, while the wave resistance is
only 2 45 tons as against 3*2 tons. At 14 knots there is a
very remarkable result in the broader ship with its fine lines,
all entrance and run, and no parallel middle body. At 14
knots the skin resistance is 8 tons as against 6*6 tons in A,
while the wave resistance is only 315 tons as against 615
tons in A.
" For a torpedo boat, 125 feet long and 51 tons displacement,
at 20 knots the skin resistance was i'2 tons, and the wave resist
ance i' i tons."
Resistance of a Completely Submerged Body. The condi
tions in this case are completely different from those which
have to be considered in the case of a vessel moving on the
surface. In this latter case waves are produced on the surface,
as we have seen, but with a completely submerged body this is
not so, provided the vessel is immersed sufficiently. We get the
clue to the form of least resistance in the shape of fishes, in which
the head or forward end is comparatively blunt, while the rear
portion tapers off very fine. The reason for the small resistances
of forms of this sort is seen when we consider the paths the particles
of water follow when flowing past. These paths are termed the
streamlines for the particular form. It will be seen that no eddies
are produced for a fishshaped form, and, as we saw on p. 306,
it is the rear end which must be fined off in order to reduce eddy
making to a minimum. This was always insisted on very strongly
by the late Mr. Froude, who said, " It is blunt tails rather than
blunt noses that cause eddies." A very good illustration of the
above is seen in the form that is given to the section of shaft
brackets in twinscrew vessels. Such sections are given in Figs.
85 E and 1 08. It will be noticed that the forward end is com
paratively olunt, while the after end is fined off to a small radius.
Horsepower, Effective and Indicated, etc. 3 1 5
Speed Coefficients. The method which is most largely
employed for determining the I.H.P. required to drive a vessel
at a certain speed is by using coefficients obtained from the
results of trials of existing vessels. They are based upon
assumptions which should always be carefully borne in mind
when applying them in actual practice.
i. Displacement Coefficient. We have seen that for speeds
at which wavemaking resistance is not experienced, the resist
ance may be taken as varying
(a) With the area of wetted surface ;
(b) Approximately as the square of the speed ;
so that we may write for the resistance in pounds
R = K a SV 2
V being the speed in knots, S the area of wetted surface in
square feet, and Kj being a coefficient depending on a number
of conditions which we have already discussed in dealing with
resistance.
Now, E.H.P =  , as we have already seen
(p. 297). Therefore we may say
E.H.P. =
where K 2 is another coefficient, which may be readily obtained
from the previous one. If now we assume that the total I.H.P.
bears a constant ratio to the E.H.P., or, in other words, the
propulsive coefficient remains the same, we may write
I.H.P. = K 3 SV 3
K 3 being another new coefficient. S, the area of the wetted
surface, is proportional to the product of the length and girth to
the waterline ; W, the displacement, is proportional to the pro
duct of the length, breadth, and draught. Thus W may be said
to be proportional to the cube of the linear dimensions, while S
is proportional to the square of the linear dimensions. Take a
vessel A, of twice the length, breadth, and draught, of another
vessel B, with every linear dimension twice that of the corre
sponding measurement in B. Then the forms of the two vessels
316 Theoretical Naval Architecture.
are precisely similar, and the area of the wetted surface of
A will be 2 a = 4 times the area of the wetted surface of B, and
the displacement of A will be 2 3 = 8 times the displacement of
B. The ratio of the linear dimensions will be the cube root
of the ratio of the displacements, in the above case ^8=2.
The ratio of corresponding areas will be the square of the cube
root of the ratio of the displacements, in the above case
(4/8) 2 = 4. This may also be written 8*. We may accord
ingly say that for similar ships the area of the wetted surface
will be proportional to the twothirds power of the displace
ment, or W'. We can now write our formula for the indicated
horsepower
W* x V 1
I.H.P. = ILJ^L
where W = the displacement in tons ;
V = the speed in knots ;
C = a coefficient termed the displacement coefficient*
If a ship is tried on the measured mile at a known displace
ment, and the I.H.P. and speed are measured, the value of the
W* X V s
coefficient C can be determined, for C = j TT p . It is usual
to calculate this coefficient for every ship that goes on trial, and
to record it for future reference, together with all the particulars
of the ship and the conditions under which she was tried. It
is a very tedious calculation to work out the term W z , which
means that the square of the displacement in tons is calculated,
and the cube root of the result found. It is usual to perform
the work by the aid of logarithms. A specimen calculation is
given here :
The Himalaya on trial displaced 4375 tons, and an I.H.P.
of 2338 was recorded, giving a speed of 12*93 knots. Find the
" displacement coefficient " of speed.
Here we have W = 4375
V = 1293
I.H.P.= 2338
1 The coefficients are often termed " Admiralty constants," but it will
be seen later that they are not at all constant for different speeds of the
same vessel.
Horsepower ) Effective and Indicated, etc. 317
By reference to a table of logarithms, we find
log 4375 = 3' 6 4io
log 1293 = i'in6
log 2338 = 33689
so that log (4375) 1 = I log 4375 = 2 '4273
log (i2'93) 3 = 3 log 1293 = 3'3348
= 2 ' 4273 + 3 ' 3348 ~
= 23932
The number of which this is the logarithm is 2473,
accordingly this is the value of the coefficient required.
2. The other coefficient employed is the " midshipsection
coefficient" 1 If M is the area of the immersed midship section
in square feet, the value of this coefficient is
M X V s
I.H.P.
This was originally based on the assumption that the
resistance of the ship might be regarded as due to the forcing
away of a volume of water whose section is that of the im
mersed midship section of the ship. This assumption is not
compatible with the modern theories of resistance of ships, and
the formula can only be true in so far as the immersed midship
section is proportional to the wetted surface.
In obtaining the W* coefficient, we have assumed that the
wetted surface of the ships we are comparing will vary as the
twothirds power of the displacement ; but this will not be true
if the ships are not similar in all respects. However, it is
found that the proportion to the area of the wetted surface is
much more nearly obtained by using W* than by using the
area of the immersed midship section. We can easily imagine
two ships of the same breadth and mean draught and similar
form of midship section whose displacement and area of wetted
surface are very different, owing to different lengths and forms.
We therefore see that, in applying these formulae, we must take
care that the forms and proportions of the ships are at any rate
somewhat similar. There is one other point about these
1 See note on p. 316.
318 Theoretical Naval Architecture.
formulae, and that is, that the performances of two ships can
only be fairly compared at " corresponding speeds." 1
Summing up the conditions under which these two formulae
should be employed, we have
(1) The resistance is proportional to the square of the speed.
(2) The resistance is proportional to the area of wetted
surface, and this area is assumed to vary as the twothirds power
of the displacement, or as the area of the immersed midship
section. Consequently, the ships we compare should be of
somewhat similar type and form.
(3) The coefficient of performance of . the machinery is
assumed to be the same. The ships we compare are supposed
to be fitted with the same type of engine, working with the
same efficiency. Accordingly we cannot fairly compare a
screw steamer with a paddle steamer, since the efficiency of
working may be very different.
(4) The conditions of the surfaces must be the same in
the two ships. It is evident that a greater I.H.P. would be
required for a given speed if the ship's bottom were foul than
if it had been newly painted, and consequently the coefficient
would have smaller values.
(5) Strictly speaking, the coefficients should only be com
pared for " corresponding speeds." 2
With proper care these formulae may be made to give
valuable assistance in determining power or speed for a new
design, but they must be carefully used, and their limitations
thoroughly appreciated. A good method of recording these
y
coefficients is to plot them on base of ^. In this way the
v L
size of ship is eliminated.
We have seen that it is only for moderate speeds that th.e
resistance can be said to be proportional to the square of the
speed, the resistance varying at a higher power as the speed
increases. Also that the propulsive coefficient is higher at the
maximum speed than at the lower speeds. So if we try a
vessel at various speeds, we cannot expect the speed coefficients
to remain constant, because the suppositions on which they are
1 See p. 319. * Seep. 319.
Horsepower, Effective and Indicated, etc. 3 ] 9
based are not fulfilled at all speeds. This is found to be the
case, as is seen by the following particulars of the trials of
H.M.S. Iris. The displacement being 3290 tons, and the area
of the immersed midship section being 700 square feet, the
measuredmile trials gave the following results :
* I.H.P. Speed in knots.
7556 186
3958 1575
1765 125
596 83
The values of the speed coefficients calculated from the
above are
Displacement Mid. sec.
coefficients. coefficients.
186 knots ... 188 ... 595
1575 , 2l8 . 690
125 243 ... 770
8'3 214 ... 677
It will be noticed that both these coefficients attain their
maximum values at about 12 knots for this ship, their value
being less for higher and lower speeds. We may explain this
by pointing out
(1) At high speeds, although the "propulsive coefficient"
is high, yet the resistance varies at a greater rate than the
square of the speed, and
(2) At low speeds, although the resistance varies nearly as
the square of the speed, yet the efficiency of the mechanism is
mot at its highest value.
Corresponding Speeds. We have frequently had to use
the terms " low speeds" and "high speeds" as applied to certain
ships, but these terms are strictly relative. What would be a
high speed for one vessel might very well be a low speed for
another. The first general idea that we have is that the speed
depends in some way on the length. Fifteen knots would be
a high speed for a ship 150 feet long, but it would be quite a
moderate speed for a ship 500 feet long. In trying a model
of a ship in order to determine its resistance, it is obvious that
we cannot run the model at the same speed as the ship ; but
there must be a speed of the model "corresponding" to the
speed of the ship. The law that we must employ is as follows :
" In comparing similar ships with one another, or ships with
320 Theoretical Naval Architecture.
models, tfo speeds must be proportional to the square root of their
linear dimensions" Thus, suppose a ship is 300 feet long, and
has to be driven at a speed of 20 knots ; we make a model of
this ship which is 6' 3" long. Then the ratio of their linear
dimensions is
300
6^5 = 4 *
and the speed of the model corresponding to 20 knots of the
ship is
20 4 \/48 = 288 knots
Speeds obtained in this way are termed " corresponding speeds"
Example. A model of a ship of 2000 tons displacement is constructed
on the \ inch = I foot scale, and is towed at a speed of 3 knots. What
speed of the ship does this correspond to ?
Although here the actual dimensions are not given, yet the ratio of the
linear dimensions is given, viz. I : 48. Therefore the speed of the ship
corresponding to 3 knots of the model is
3 \/48 = 20^ knots
Expressing this law in a formula, we may say
where V = speed in knots ;
L = the length in feet ;
c = a coefficient expressing the ratio V : \/L } and
consequently giving a measure of the speed.
We may take the following as average values of the co
efficient " c" in fullsized ships :
When c = 0*5 to 0*65, the ship is being driven at a
moderate economical speed ;
c = 07 to 1*0, gives the speed of mail steamers and
modern battleships ;
c = i'o to i '3, gives the speed of cruisers.
Beyond this we cannot go in fullsized vessels, since it is not
possible to get in enough enginepower. This can, however, be
done in torpedoboats and torpedoboat destroyers, and here we
have c = 19 to 25. These may be termed excessive speeds.
The remarks already made as to wave resistance gives the
reason for the above. For low speeds the wavemaking resist
ance is small. When, however, the speed increases such that
the length of the wave is about the length of ship, we have the
Horsepower, Effective and Indicated, etc. 321
maximum interference, and the rate of increase of resistance
with increase of speed is greatest. If V is speed in knots, the
V 2
length of accompanying wave is ; when the wave equals the
I'o
V
length of ship, we have =  133. So that when the ratio
V L
V
j= is unity and somewhat above, the resistance is increasing
v L
very rapidly. If the speed can be pressed beyond the above,
we reach a state of things where the wave is longer than the
length of boat, and although the resistance is very high yet it is
not increasing at so great a rate. This can only be the case in
vessels of the destroyer or motor type. The following figures
show how the total resistance varies in a typical destroyer :
Up to ii knots as second power nearly, at 1 6 knots as V 3 ,
from 1 8 to 20 knots as (V) 33 , at 22 knots as (V) 2 ' 7 , at 25 knots
as V 2 , and at 30 knots as V 2 nearly. The maximum rate of
increase is at 1 8 to 20 knots, and here the accompanying wave
approximated to the length of the ship.
Froude's Law of Comparison. This law enables us
to compare the resistance of a ship with that of her model, or
the resistances of two ships of different size but of the same
form. It is as follows
If the linear dimensions of a vessel be I times the dimensions
of the model, and the resistance of the latter at speeds V 1} V 2 , V 3 ,
etc., are R 15 Rg, R 3 , etc., then at the ''corresponding speeds" of
the ship, Vj^T V 2A/Z V 8 V"Z etc., the resistance of the ship will
be Ri/ 3 , R 2 / 3 , R 3 / 3 , etc.
In passing from a model to a fullsized ship there is a
correction to be made, because of the different effect of the
friction of the water on the longer surface. The law of com
parison strictly applies to the resistances other than frictional.
The law can be used in comparing the resistance of two
ships of similar form, and is found of great value when model
experiments are not available.
In the earlier portion of this chapter we referred to the
experiments of the Greyhound by the late Mr. Froude. A
curve of resistance of the ship in pounds on a base of speed
Y
322
Theoretical Naval Architecture.
is given by A, in Fig. 106. In connection with these experi
ments, a model of the Greyhound was made and tried in the
experimental tank under similar conditions of draught as the
ship, and between speeds corresponding to those at which the
ship herself had been towed. The resistance of the model having
been found at a number of speeds, it was possible to construct
a curve of resistance on a base of speed as shown by C in
Fig. 113. The scale of the model was ^ full size,_jmd
therefore the corresponding speeds of the ship were V "16, or
four times the speed of the model. If the law of comparison
125
SPEED
FIG. 113.
held good for the total resistance, the resistance of the ship
should have been i6 3 = 4096 times the resistance of the model
at corresponding speeds ; but this was not the case, owing to
the different effect of surface friction on the long and short
surfaces. The necessary correction was made as follows.
The wetted surface of the model was calculated, and by
employing a coefficient suitable to the length of the model and
the condition of its surface, the resistance due to surface
friction was calculated for various speeds as explained (p. 305),
and a curve drawn through all the spots thus obtained. This
Horsepower, Effective and Indicated, etc. 323
is shown by the dotted curve DD in Fig. 113. Thus at
250 feet per minute the total resistance of the model is given
by ac, and the resistance due to surface friction by ad. The
portion 'of the ordinate between the curves CC and DD will
give at any speed the resistance due to other causes than that
of surface friction. Thus at 250 feet per minute, these other
resistances are given by cd. This figure shows very clearly
how the resistance at low speeds is almost wholly due to
surface friction, and this forms at high speeds a large propor
tion of the total. The wavemaking resistance, as we have
already seen, is the chief cause of the difference between the
curves CC and DD, which difference becomes greater as the
speed increases. It is the resistance, other than frictional, to
which the law of comparison is intended to apply.
We have in Fig. 106 the curve of resistance, AA, of the
Greyhound on a base of speed, and in precisely the same way
as for the model a curve of frictional resistance was drawn in
for the ship, taking the coefficient proper for the state of the
surface of the ship and its length. Such a curve is given by
BB, Fig. 1 06. Then it was found that the ordinates between
the curves A A and BB, Fig. 106, giving the resistance for the
ship other than frictional, were in practical agreement with the
ordinates between the curves CC and DD, Fig. 113, giving
the resistance of the model other than frictional, allowing for
the " law of comparison " above stated. That is, at speeds of the
ship V 1 ^, or four times the speeds of the model, the resistance
of the ship other than frictional was practically i6 3 , or 4096
times the resistance of the model.
These experiments of the Greyhound and her model form
the first experimental verification of the law of comparison.
In 1883 some towing trials were made on a torpedoboat
by Mr. Yarrow, and a model of the boat was tried at the
experimental tank belonging to the British Admiralty. In this
case also there was virtual agreement between the boat and
the model according to the law of comparison. It is now the
practice of the British Admiralty and others to have models
made and run in a tank. The data obtained are of great
value in determining the power and speed of new designs.
3 2 4 Theoretical Naval Architecture.
For further particulars the student is referred to the sources
of information mentioned at the end of the book.
Having the resistance of a ship at any given speed, we can
at once determine the E.H.P. at that speed (see p. 297), and
then by using a suitable propulsive coefficient, we may deter
mine the I.H.P. at that speed. Thus, if at 10 knots the resist
ance of a ship is 10,700 Ibs., we can obtain the E.H.P. as
follows :
Speed in feet per minute = 10 x
Work done per minute = 10,700 x (10 x ^p) footlbs.
10700 x
E.H.P. =
33000
328
and if we assume a propulsive coefficient of 45 per cent.
X zoo
45
= 729
By the use of the law of comparison, we can pass from one
ship whose trials have been recorded to another ship of the
same form, whose I.H.P. at a certain speed is required. It is
found very useful when data as to I.H.P. and speed of existing
ships are available. In using the law we make the following
assumptions, which are all reasonable ones to make.
(1) The correction for surface friction in passing from one
ship to another of different length is unnecessary.
(2) The condition of the surfaces of the two vessels are
assumed to be the same.
(3) The efficiency of the machinery, propellers, etc., is
assumed the same in both cases, so that we can use I.H.P.
instead of E.H.P.
The method of using the law will be best illustrated by the
following example :
A vessel of 3290 tons has an I.H.P. of 250x3 on trial at 14 knots. What
would be the probable I.H.P. of a vessel of the same form, but of three
times the displacement, at the corresponding speed ?
Horsepower, Effective and Indicated, etc. 325
The ratio of the displacement = 3
/. the ratio of the linear dimensions / = v/3
= i '44
.*. the corresponding speed = 14 X v I '44
= i6'8 knots
The resistance of the new ship will be /* times that of the original, and
accordingly the E.H.P., and therefore the I.H.P., will be that of the
original ship multiplied by / = (1*44)* = 3*6, and
I.H.P. for new ship = 2500 X 3*6
9000
When ships have been run on the measured mile at pro
gressive speeds, the information obtained is found to be ex
tremely Valuable, since we can draw for the ship thus tried a
curve of I.H.P. on a base of speed, and thus at intermediate
speeds we can determine the I.H.P. necessary. The following
example will show how such a curve is found useful in
estimating I.H.P. for a new design :
A vessel of 9000 tons is being designed, and it is desired to obtain a
speed of 21 knots. A ship of 7390 tons of similar form has been tried, and
a curve of I.H.P. to a base of speed drawn. At speeds of 10, 14, 18, and
20 knots the I.H.P. is 1000, 3000, 7500, 11,000 respectively.
Now, the corresponding speeds of the ships will vary as the square root
of the ratio of linear dimension /.
We have
/*?
and / = i "07
. V*"= 1035
therefore the corresponding speed of the 739Oton ship is
21 T i '035 = 20*3
By drawing in the curve of I.H.P. and continuing it beyond the 20
knots, we find that the I.H.P. corresponding to a speed of 20*3 knots is
about 1 1,700. The I.H.P. for the goooton ship at 21 knots is accordingly
11,700 x / = 11,700 x 1*26
= 14,750 I.H.P. about
PROOF OF THE LAW OF COMPARISON.
Take the following symbols :
P for force.
m for mass.
/ for acceleration.
/ for time.
v for velocity.
/ for length.
326 Theoretical Naval Architecture.
Then force P = (mass X acceleration)
= ( x/)
velocity = (/ 5 /)
Acceleration is increase of velocity in unit time, = (/:
Mass varies as the volume or P.
Force, which equals (in X/), may be written
/. if j is constant, force will vary as / 3 .
Progressive Speed Trials, It is now the usual practice
to run vessels at a series of speeds from a low speed up to the
highest speed attainable in order to construct a curve of power,
etc., on base of speed. Such a record is of the highest value
as data for design purposes, and the information obtained as to
slip of propellers will frequently indicate the direction in which
improvements may be made. At each speed it is necessary to
obtain simultaneously the revolutions, I.H.P., 1 and speed. The
usual practice is to run the ship on a measuredmile course.
Fig. H3A shows such a course. Two pairs of posts, AB and
 =.IJ<HfiTL=7
I SMIPS_ _CouRSg
Fie. i ISA.
1 Indicator cards are taken from each piston, showing how the pressure
of the steam varies at each point of a revolution. A calculation from
these cards enables the I.H.P. to be determined. For turbine engines no
corresponding method is available. A method of determining power of
turbine machinery has been introduced by Mr. Johnson of Messrs. Denny
Bros., Dumbarton, by measuring the torsion of the shaft by electrical
instruments. Another method was described by Mr. Gibson of Messrs.
Cammell Laird (see I.N.A. for 1907). For a general discussion of the
subject see N.E. Coast Inst., 1908.
Horsepower, Effective and Indicated, etc. 327
CD, are placed exactly a knot (6080 feet) apart, and the
ship's course is steered at right angles. The time of transit
is taken by a chronometer stopwatch. In order to eliminate
the effect of tide, several runs are taken both with and against
the tide, and the " mean of means" is taken. Thus, suppose a
vessel has four runs, and the speeds observed are 15*13, 14*61,
1566, 1411 knots respectively. Then the "mean of means"
is obtained as follows :
First Second Mean of
Speeds, means means means
X 2. X 4 X 8.
I5 ' 13 i 20741,
1461 y/4 }6o'oi
,,\ 30*27 \ \ 120*05
*66 6 '*
,,
15*66] o04
20*77
14*113 ** I*
The true mean speed is therefore 120*05 r 8 = 15006 knots.
The ordinary mean of the speeds is 1488 knots. The same
result as the mean of means is got by multiplying by i, 3, 3, i
and dividing by 8.
The above is based on the assumption that the speed of tide can be
expressed as a quadratic function of the time. That is, if y is speed of
.ide, then
y = a + a^t f a^t*
t being the time, a , a,, a 2 being constants.
Thus, when t = o, speed of tide y l a t
t=t
'=3* J4 = o
If V is the true speed of ship, then, owing to the tide, the speed at intervals
of / up and down the mile will be
(V + j,), (V ^), (V +j 3 ), (V y.)
or a mean of means of
V + 1(3^3 y*y*yi)
By substituting in the above values for y lt etc., this is seen to be equal to V.
If six runs are taken up and down, the mean of means is obtained by
multiplying by I, 5, 10, 10, 5, I and dividing by 32, and it is easily shown
that if the tide be assumed a cubic function of the time, the " mean of
means " at equal intervals of time gives the true mean speed.
It is necessary to run measuredmile trials in deep water,
or a falling off in speed will be experienced. If the water is
not deep, the natural streamlines are not formed round the
ship, and this restriction is a serious cause of resistance. A
328
Theoretical Naval Architecture.
similar thing is noticed in canals. A conspicuous instance
was noticed on the trials of H.M.S. Edgar. When tried at
Stoke's Bay, with a depth of water of 12 fathoms, 13,260 horse
power was required for 2oJ knots. On the deepsea course
between Plymouth and Falmouth, 2 1 knots was obtained with
12,550 horsepower, or about f knot difference for the same
power. In consequence of this, trials at high speeds must be
carried out on a deepwater course, the finest probably being
at Skelmorlie, near the Clyde, where the depth of water is 40
fathoms.
Colonel English's Experimental Method of deter
mining I.H.P. of a New Design by the Use of Models
(I.M.E., 1896). This method of determining the power for
a new design is an interesting application of the principles of
the present chapter.
Two models are made, one of a known ship, the other of
the new design, on such scales that when towed at the same
speed they shall be at the corresponding speeds proper for each.
In the following table the capitals refer to the ships, and the
small letters to the models, and the resistance is divided into
the frictionai and wavemaking. It will be remembered that
the law of comparison only strictly holds for wavemaking
resistance.
Resistance.
Sn^^H
Frictionai.
Wavemaking.
Actual ship (i) ...
F,
w,
D,
v.
New design (2)
F 2
W 2
? 2
v f
Model of (I)
/.
w l
4
v \
Model of (2)
7*
t
d,
Vt
By the law of comparison
and if the models are towed at the same speed, v^ z/ 2 ; so
that
Horsepower, Effective and Indicated, etc. 329
4 D 2 A Vi
This determines the relative scale of models, and
The total resistance of model (i) = / + w^ and that of model
(2) =/ 2 + a/ a . Let/ 2 + a> a = (/ + Wj), say.
The law of comparison indicates that the wavemaking
resistance varies as the displacement, so that
w, D,
so that w, = * .
. + n ./ / a
We want to get the wavemaking resistance of the new design,
viz. W 2 ; we first find w. 2 from the above, and we can calculate
/[ and/ 2 by the use of appropriate frictional coefficients. To
get W 1} we proceed as follows : For the known ship we have
data regarding I.H.P. at speed V tl this can be turned into
E.H.P. by the use of a propulsive coefficient, and this E.H.P.
FIG. 1136.
can be turned at once into resistance, which is (F x + Wj). The
frictional resistance can be calculated by the ordinary rules,
and we have left W 1} the wavemaking resistance of the known
ship. The only part of the above expression we do not know
330 Theoretical Naval Architecture.
is n. This is obtained by towing the models abreast of one
another and adjusting so that they are exactly abreast (Fig.
1133). When this is so, the ratio of the levers determines the
ratio n. We thus can determine a/ 2 , and W 2 = w 2 . 2 . F 2 can
a 2
be calculated, so that F 2 + W 2 is determined. This is turned
into E.H.P. at the speed V 2 , and, using the same propulsive
coefficient as before, the I.H.P. is found for the new design
at speed V 2 . The models were of yellow pine ballasted to
desired draught. A small electric motor was used for towing,
and when the levers were adjusted so that the models towed
abreast, the only measurement necessary was the ratio between
the levers.
The method may be made clearer by reference to an
example. It is desired to know the I.H.P. to drive a destroyer
of 300 tons displacement at a speed of 30 knots, and a known
destroyer of 247 tons required 3915 I.H.P. for a speed of
27*85 knots. The model of this vessel was made on a scale
of ^, so that the speed corresponding to 2785 knots was
, =6*23 knots. The scale of the model of the new ship
v 20
must be such that 623 knots of model corresponds to 30 knots
of the ship, giving a scale of (  H \ = _L
The wetted surface of known ship was calculated to be
3796 square feet, so that that of model was 3796 x (^j) 2 = 9*5
The wetted surface of new ship was 4321 square feet, and of
(I \ 2
 ) = 802. Using these values and
appropriate values for the coefficient of friction, we have
F! (known ship) = 00094 x 379 6 X (27'85) 183 = 15,720 Ibs.
/ (its model) = 001124 x 9*5 X (6'23) 1>85 = 315 Ibs.
F 2 (new ship) = 0*0094 X 4321 X (3o) 1 ' 83 = 20,500 Ibs.
/ a (its model) = 001124 X 8*02 x (6'23) 1W = 2*66 Ibs.
The propulsive coefficient being assumed as o'6, we have
E.H.P. of known ship o'6 X 3915 = 2349, so that the total
resistance of ship was
Horsepower, Effective and Indicated, etc. 331
therefore Wj = 11,747 Ibs.
From the towing trial at 6*23 knots, n = 0*811, so that
a' 2 = o'8n. (^) 3 . 11,747 + ' 811 X 3*15  2'66 == ro8 Ibs.
we therefore have
W 2 = ro8 x (232)* = 13,500 Ibs.
The total resistance of new ship is therefore 34,000 Ibs., and
assuming the same propulsive coefficient, we have
I.H.P. = 3^ X 34,000 X 30 X ^ = 5220
Calculation of E.H.P. Mr. A. W. Johns gave before
the I.N.A., 1907, a table which gives the E.H.P. due to skin
friction for a number of speeds and lengths of ship, based on
Mr. Froude's constants and on the assumption that the skin
friction varies as V 1825 .
If S is the wetted surface in square feet, then E.H.P. =/. S,
where /"has the values given in table, p. 332.
In the same paper he gave a series of curves based on
model experiments, reproduced in Fig. 1130, from which,
knowing the prismatic coefficient of fineness, the residuary
horsepower can be obtained. The curves are drawn for a
/ V \ a V 2
number of values of ( ^\ = (where L is underwater
length) varying from 05 to 1*3 on a base of prismatic co
efficients varying from 0*52 to 0*74. It is very striking to
note how rapidly the residuary horsepower increases, for high
values of speedlength ratio, with increase of prismatic co
efficient. The prismatic coefficient has been taken with the
length P.P., and Mr. Johns states that for merchant ships
better results are obtained by increasing the prismatic coefficient
by 0*02, this being due to the fact that in such vessels the
length P.P. is practically the immersed length of the ship, and
not, as in the majority of warships, an appreciably smaller
length. In a few ships of exceptionally good form the curves
give too great a result, but for ordinary forms of ships the
curves give a good approximation to the results obtained from
332
Theoretical Naval Architecture.
FRICTIONAL RESISTANCE PER SQUARE FOOT OF WETTED SURFACE.
Speed
in
knots.
Length of ship in feet.
TOO
150
200
300
400
500
600
800
1000
IO
00188
00186
0*0184
00183
00181
00180
00179
00176
0*0174
II
00246
00243
0024I
00239
00237
00235
0*0233
00231
00228
12
00315
0*0312
00309
00307
00304
00302
00300
00296
00293
'3
00397
00390
00387
00384
00381
00378
00375
00371
00367
14
00489
00451
00478
00473
00469
00466
00463
00458
00453
15
00594
00585
00580
00575
00570
0*0567
00563
00557
00551
16
00713
OO7O2
00697
00690
00685
0*0680
0*0675
0*0668
0066 1
17
0*0846
00833
0*0827
00819
00812
00807
00802
0*0793
00784
18
00995
00979
00972
00962
00955
00948
00942
0*0931
00921
*9
0II59
0*114!
0II32
OII2I
OIII2
01105
0*1098
o 1086
01074
20
01340
0*1319
01308
0*I296
01286
01277
01268
01254
01241
21
01537
01514
0I502
0I487
0I476
01466
0*1456
0*1440
0*1424
22
01753
0*1726
01713
01696
0I683
0*1672
0*1661
01643
01625
2 3
01988
01957
0*1942
0I923
0I908
01895
01882
01861
01841
24
02242
02207
0*2190
0*2l69
02I52
0*2138
02124
02IOI
0*2078
25
02516
0*2477
0*2458
02434
024I5
02399
02383
02357
0*2331
26
028II
0*2767
0*2746
027I9
02698
02680
02662
02633
0*2604
27
03126
0*3078
03054
03025
0300I
02981
0*2961
02929
02897
28
03466
034I2
0*3386
03353
03327
03305
03283
03247
03211
29
03826
03767
03738
03702
03673
03649
03624
03585
Q3545
30
04210
04I45
04113
04073
0404!
0*4014
03988
03944
0*3900
31
0*4624
04552
045 '7
04473
04438
0*4409
04379
04322
04274
3 2
0*5050
04972
04934
04886
04848
04816
04784
04732
0^4680
33
Q'5499
054I4
05372
05320
05278
05243
05208
05I5 1
0*5094
34
05995
05902
05857
0*5800
05755
05707
0*5679
05617
o'5555
35
06508
06407
0*6358
06296
06247
06206
06164
06097
06030
36
07047
06938
06885
06818
06765
06720
06675
06603
06530
37
07611
07494
07436
07364
o73>6
07258
07209
07130
07051
38
0*8209
08082
0*8020
07942
07880
07828
07776
07691
07606
39
08835
08698
08631
08547
08480
08424
08356
0*8276
08185
40
09490
09343
09271
09181
09109
09049
08989
0*8890
08792
NOTE. The above table has been extended beyond that given in Mr. Johns'
paper to include lengths of 1000 feet and speeds up to 40 knots.
Horsepower p , Effective and Indicated, etc. 333
model experiments. The curves apply to vessels in which the
ratio beam/draught varies from about 27 to 2*9. For greater
ratios than the latter the curves give results which are smaller
than they should be, whilst for smaller ratios than the former
the results will be too great.
As an example, take a vessel 500 ft. (P.P.) X 71 ft. X 26ft.
X 14,100 tons, prismatic coefficient 0582. Underwater length
520 ft.
334
Theoretical Naval Architecture.
The approximate wetted surface by Denny's formula
17 L.D + = i"j X 500 X 26 +
14,100 x 35
26
= 41,100 square feet
and by Taylor's formula
J 5'5 Vw.L = 41,200 square feet
Taking 42,000 and using the coefficients in table, we obtain
the following values of E.H.P. due to surface friction, from
16 to 25 knots, viz.: 2860, 3390, 3980, 4620, 5360, 6150,
7020, 7950, 8960, 10,100.
Now going to the curves and erecting an ordinate at
0*582 prismatic coefficient, the values of coefficient at speeds
161, 17*65, 191, 20*4, 2 1 '6, 228, 2395, 2 5 knots are
measured as 0015, C ' 02I J 0*029, 0*042, 0*060, 0081, .0104,
0*138, which have to be multiplied by (displacement in
tons) t, giving us 1040, 1450, 2000, 2900, 4150, 5600, 7200,
9550 residuary horsepower.
The above results, plotted as in Fig. 1130, give us an
estimated curve of total E.H.P.
To obtain the Space which must be passed over
by a Ship starting from Rest to any Speed short
of the Full Speed, supposing the Engines are
Horsepower, Effective and Indicated, etc. 335
exerting the Thrust corresponding to the Maximum
Speed. (a) Supposing the resistance is varying as the square
of the speed. When a ship is being accelerated through the
water there is a certain amount of water accompanying the
ship which has to be accelerated as well. This is usually
taken (based on the Greyhound experiments of Mr. W. Froude)
as 20 per cent, of the weight of the ship. The virtual mass
/W\
to be accelerated is therefore f . ( J, where g is the accele
ration due to gravity (32*2 in footsecond units).
Let R be the resistance of ship at full speed V.
r lower speed #.
Then the force urging the ship is the constant thrust of the
propeller = R and the force accelerating the ship is R r.
Now by the principles of dynamics
Force = mass X acceleration
or/=f.
W
Now / = acceleration = ~
__dv ds _ dv
~ ~ds'~dt~ v "ds
dv R  r
so that .=!._,_. or
W v
and on integrating
w rvi v
= s ' T } R _ r dv to speed Vj from rest.
' g
Now on the assumption that resistance varies as the square
of the speed
V ~VV
336 Theoretical Naval Architecture.
and on integrating
W V 2 / V 2
and the space from speed V x to speed V 2 is
1 e W Y! i /V 2  V 2 2 \
2 5  R 10 g<  \ V 2  V^/
() Without making any assumptions as to the variation of
resistance with speed, if we have a curve of I.H.P. on base of
speed, we can get a good approximation to the space required
to go from one speed to another short of the maximum suppos
ing the full thrust due to the top H.P. is exerted from the start.
Take as an example a vessel of 5600 tons, whose I.H.P. at
speeds of 10, 12, 14, 16, 18 and 20 knots are respectively 950,
1640, 2720, 4340, 6660 and 10,060. It is desired to obtain
the space required to increase the speed from 10 to 18 knots,
supposing the engines are exerting the full thrust corresponding
to 20 knots.
Here the virtual weight is x 5600 = 6720 tons, and
assuming I.H.P. = 2 E.H.P. all through
101 X 2240
(I.H.P.),, = 2 X r X v X Q r in tons, v in knots
(I.H.P.). i '
(I.H.P.n
V J
_
"' R ~ = 137 20
Taking v as 10, 12, 14, 16, and 18 knots respectively
i /io,o6o Q'joX
R  r at TO knots = ~ ~ ~ ~ = 2 9'8 tons
i /io,o6o i64o\
R  r at 12 knots = f ^  jyJ = 26'8 tons
I / 10, 060 2720\
R  r at 14 knots = ^(^ ) = 22< 5 tons
i /io,o6o 4UO\ ,
R  r at 1 6 knots = ^(^  ^) = l6 ' 9 tons
i /io,o6o 666o\
R  r at 18 knots = ^(fe  ^~) = 97
tons.
Horsepower, Effective and Indicated, etc. 337
ds e W v
Now =. .
dv ' g R r
ds\ 6720 6080 10
= ' I8 '5 m foot second unlts
and similarly
'ds\ , (ds\ ids\ /ds\
l = 158,! ) =220,177) = 334, ( j =657.
/ds
fa. dv, so that we can obtain the integration by
means of Simpson's first rule having values of T.
We therefore have
space from 10 to 18 knots
= \ X 3'38 . [1185 + 4(i58) + 2(220) + 4(334) + 657]
= 3600 feet
(3*38 being the equivalent in footsecond units of 2 knots, the
interval chosen).
It will have been seen in the above example that special
attention is necessary to the units which have been taken as
tons, feet and seconds.
The time taken can be obtained in a similar manner by
integrating values of ( r ) taken at equal intervals of time.
Example. The I.H.P. of a vessel of 14,200 tons at 10, 12, 14, 16, 18,
and 20 knots are 1750, 3150, 5000, 7600, 10,850 and 15,300. Supposing
the vessel is exerting 15,300 LH.P., how far would the ship travel in
going from 10 to 18 knots, and how long would it take.
Ans. 7075 feet. 282 seconds.
PROPULSION.
The following notes have been prepared in order to
provide an introduction to the subject. The subject is too
large to be dealt with adequately in the space at disposal, and
for fuller information reference must be made to the systematic
treatises given at the end of the book.
Wake. We have dealt above with the various resistances
z
338 Theoretical Naval Architecture.
which oppose a vessel's progress through the water. These
are mainly frictional and wave making. The friction of the
water on the surface of the vessel is the cause of a surrounding
zone of water following in the direction of motion and the
forward velocity of this zone increases as we go aft. The
consequence is that at the stern there is a belt of water having
a forward velocity. This velocity is variable in amount and
in direction, but may be assumed, in the case of each propeller,
to have the same effect as a body of water having a certain
uniform velocity forwards. This body of water is termed the
frictional wake. The speed of the wake is conveniently
expressed as a fraction of the speed of the ship, say x . V.
The wake will have a higher velocity nearer the middle line
of the ship than at points farther away. The importance of
this wake is due to the fact that the propeller has thus to work
in water which has this forward velocity, and therefore the
speed of the propeller through the water is not the speed of the
ship V, but (i *)V = V 1} say. 1 The propeller derives
increased thrust from this cause, and a single screw will benefit
more than twin screws, owing to the fact above mentioned as
to the greater velocity of the water nearer the middle line.
The frictional wake is caused by the motion of the ship, and
the increased thrust may therefore be regarded as the return
of a small portion of the energy spent by the ship in over
coming the friction of the water on the surface.
The simple frictional wake above described is complicated
by the existence of other factors, viz. :
(a) The stream line wake. (We have seen that the
stream lines closing round a ship tend to a
diminution of velocity and an increase of
pressure.)
(b) The presence of a wave at the stern. (If the crest
of a wave is over the propeller the particles
1 In Froude's notation the speed of wake is expressed as a fraction of
Vj, say w . V It so that speed of propeller through the water is V w/V^ or
 t = i +w, w being called the "wake percentage." It follows that
Horsepower, Effective and Indicated, etc. 339
of water in their orbital motion are moving
forwards. If there is a trough the particles are
moving backwards.)
The information as to the value of the speed of the wake
is scanty, and in systematic propeller design it is necessary to
assume some value. Mr. R. E. Froude assumed 10 per cent,
of the velocity of the ship as a standard value for the wake,
i.e. x = 0*1. The following formulae have been obtained as
the results of Mr. Luke's investigations (I.N.A. 1910) :
Twin screws x 02 f 0*55 (block coeff.).
Single screws x = 005 4 05 (block coeff.).
The ratio V x 4 V represents what may be termed the wake
, V(i  x)
gam factor, and this is  5 r=  = i x.
Augmentation of Resistance. Anything which inter
feres with the natural closing in of the stream lines at the
stern of a ship will cause an increase of resistance. The
presence of the propeller at the stern is such an interference,
and gives rise to an augment of resistance. This will be
greater in a single screw than in a twin screw ship, since the
propellers in the latter case are further away from the middle
line of the ship. It is thus seen that, although a single screw
ship stands to gain more from the frictional wake than a
twin screw ship, yet it loses more from the augment of
resistance.
Thrust Deduction. Instead of regarding this loss as an
augment of resistance, it is preferable to regard it as a loss in
the thrust of the propeller. If T be the thrust required to
overcome the resistance of the ship plus the augment, and R
the thrust required to overcome the resistance only, then
T  R is termed the thrust deduction, and T  R = / . T, so
that R = T(i  /), and (i  t) is called the thrust deduction
factor.
Hull Efficiency. The useful work done by the ship is
the product of the resistance and the speed, or R x V. The
work done by the propeller is the product of the thrust and the
speed the propeller passes through the water, or T x V lt
340
Theoretical Naval Architecture.
The ratio (R X V) ^ (T x VJ is termed the hull efficiency,
and may be written from the above . The usual value
i x
assumed for this is unity, the gain due to the wake being
balanced by the increase of thrust due to the augment of
resistance.
The following table, taken from Prof. Dunkerley's recent
book on " Resistance and Propulsion," expresses admirably
the way the work available in the engines is expended, and
what portion of the work is lost beyond recall. The notation
is that employed in the present notes.
Work of engines in
uniform motion
[I.H.P.].
Loss of work
in friction.
Work delivered to propeller
by the shaft [S.H.P.].
I
Available w
wake = [T(
We
me
ork from Available work due
V V)]. to engines
[P.H.P.JlTxVJ.
Loss of work at pro
peller due to shock,
friction, and causing
useless motions.
I
Total available work
[T.H.P.] = [T X V].
rk necessary to Towrope work
vercome aug [E.H.P.] = [R x V].
nted resistance
[(T  R)V]. 1
Work available
in wake.
1
Work lost beyond
recall.
Horse Power. We have seen that the horse power
necessary to overcome the towrope resistance of a ship
at a given speed is the Effective HorsePower^ E.H.P.
Leaving out the constants, E.H.P. = R x V. The thrust
K H P
horsepower, U T.H.P. = T . V =  ^~. The propeller
Horsepower , Effective and Indicated, etc. 341
horsepower , P.H.P., makes allowance for the gain due to the
wake = T X Y! = T . V(i  #), so that
P.H.P. = T.H.P.(i  x) and P.H.P. = E.H.P. ^^,
i.e. E.H.P. = P.H.P. x hull efficiency.
There are certain losses in the propeller due to the
frictional and edgewise resistance of the blades and to the
rotary motion imparted to the water. The ratio between
the P.H.P. and the shaft horsepower, S.H.P., is the measure
of the efficiency of the propeller, or P.H.P. 4 S.H.P. = e.
In reciprocating engines the power actually exerted in the
cylinders, or I.H.P., is greater than the S.H.P., the relation
between the two being the measure of the efficiency of the
machinery, or S.H.P. r I.H.P. = e m .
The ratio E.H.P. 4 I.H.P. is the propulsive coefficient,
and tracing through the various stages,
E.H.P. _ E.H.P. T.H.P. P.H.P. S.H.P.
I.H.P. ~ T JLP7 X "PlLP; X S.H.P. X I.H.P.
=   X X e X e m .
i i x
Taking a case in which the engine efficiency e m = 0*85,
propeller efficiency e = 0*65, x and / each 0*15, the propulsive
. 085 i
coefficient is 4 x ^~ X 0*65 x 085 = 55^25 per cent,
i 0*05
With turbine machinery it is usual to assume that the
propulsive coefficient is the ratio between the E.H.P. and the
horsepower being transmitted through the shaft inside the
ship. Owing to the high revolutions at which turbines work,
the propellers connected directly to them have a low efficiency,
and it is found (as e.g. in the Lusitania) that the propulsive
efficiency thus denned is about 50 per cent. 1
Cavitation. The force which pushes the ship along is the
reaction from the projection in a sternward direction of the
water by the propeller. The momentum of this water, per
unit of time, is the measure of the thrust which is transmitted
to the ship through the thrust block. The water will not
1 The use of the geared turbine enables great allround efficiency to be
obtained, as the turbine can run fast and the propeller run slow.
342 Theoretical Naval Architecture.
follow up at the back of the blades of the propeller if the
thrust is too great and if the velocity of the blades is sufficiently
high. This causes a loss of thrustproducing power, and is
termed cavitation. Mr. Speakman (Scottish Inst. E. and S.,
1905), from an analysis of numerous trials, considers that to
avoid cavitation the limit of pressure per square inch of
projected surface should be about i Ib. for every 1000 feet of
circumferential velocity of blade tips. Mr. Sidney Barnaby
assigns nj Ibs. per square inch as the maximum average
thrust per square inch of projected area. These are for an
immersion of tip of 12 inches; for each additional foot of
immersion f Ib. per square inch may be added. This figure
of nj Ibs., however, may be exceeded for propellers with
turbine machinery, owing to the uniform turning moment. 1
The thrust of the screw is obtained as follows : T.H.P.
P TT p
= T 1 ^; / in the absence of definite information may be
taken as o'i. T.H.P = ^ T X V (V in knots, T in pounds),
T.H.P. I.H.P. .
or thrust = 326 = 181 X y , taking a propulsive
coefficient of 05. The I.H.P. is that for the screw in question.
Taking 1125 Ibs. per square inch of projected area A p , we
have
Projected area of blades 1 _ 181 I.H.P.
in square feet J ~ A * ~~ 1125 xT44 X V~
I.H.P.
= O'll X y
The relation between the developed blade area and
projected blade area for the Admiralty pattern blade 2 is given
by Mr. Barnaby as follows :
Developed area = projected area \/i f o'425(pitch ratio) 2 .
The projected blade area is often expressed as a fraction
of the disc area, and Mr. Speakman gives the following as
usual values of this ratio :
1 Mr. Baker in his book takes 13 Ibs. per square inch to obtain the
minimum (developed) area.
2 The Admiralty pattern blade when developed is an ellipse whose
major axis is the radius of the propeller and whose minor axis is the major
axis. Propellers for turbines have greater width ratio than this.
Horsepower, Effective and Indicated, etc. 343
Reciprocating machinery (naval practice)
Large ships . , , 033
Destroyers . 033 0*4
Turbines . . . . o  4 0*56
The reason for the large value in turbine vessels is the
excessive speed of rotation, which causes cavitation unless a
large area is given. The diameter is also brought as small as
possible in order to avoid excessive velocity of the blade tips,
which has, however, been as high as 14,850 feet per minute.
Pitch. A screw propeller usually has the driving face
(i.e. the after side) in the form of a true screw, and the pitch
of the screw is denned as the distance this face would advance
in one revolution if working in a solid nut. If a variable
pitch is used, then there is an equivalent pitch for the whole
surface which may be found. The speed of screw is the distance
it would advance in one minute if working in a solid nut. If
N is the revolutions per minute and P the pitch in feet, then
speed of screw is P x N feet per minute. The ratio of pitch
to diameter (P f D) is the pitch ratio /.
Slip. The advance of a screw through the water when
propelling a ship is not the speed of the ship V, because of
the presence of the wake. This speed is V x = (i #)V, where
^V is the speed of wake. The difference between the speed
of the screw as defined above and its speed forward relative to
the water in which it is working is termed the slip, or
slip = N.P.  y, Vj (Vi in knots)
N.P.^.V,
oo
The slip ratio, s = ^ p
This is the true slip, but as we do not generally know the
value of V 1} the apparent slip is usually dealt with, being the
difference between the speed of screw and the speed of ship.
N.P.^.V
6o
Apparent slip ratio s l =  j^p  (V in knots)
344 Theoretical Naval Architecture.
Y! being less than V, it follows that the real slip is greater
than the apparent slip. Cases are on record in which a
negative apparent slip has been obtained, which means that the
sternward speed of the water from the screw is less than the
forward speed of the wake. It is, of course, impossible to
have the true slip a negative quantity, as this would involve a
thrust being exerted without the projection of water in a
sternward direction.
It follows from the above that the true slip and the apparent
slip are connected by the following :
s = x + j x ( i  x).
Thus, an apparent slip ratio of 20 per cent, with a wake of
10 per cent, means a true slip of 28 per cent.
The second volume of Prof. Biles' " Design and Construc
tion of Steam Ships" deals exhaustively with the propeller
question, and gives the methods in vogue for determining
propeller dimensions based on the model experiments of
Froude and Taylor.
See also Mr. Baker's book referred to at the end of the
book.
EXAMPLES TO CHAPTER VIII.
1. The Greyhound was towed at the rate of 845 feet per minute, and
the horizontal strain on the towrope, including an estimate of the air
resistance of masts and rigging, was 6200 Ibs. Find the effective horse
power at that speed.
Ans. 159 E.H.P. nearly.
2. A vessel of 5500 tons displacement is being towed at a speed of
8 knots, and her resistance at that speed is estimated at 18,740 Ibs. What
horsepower is being transmitted through the towrope ?
Ans. 460.
3. A steamyacht has the following particulars given :
Displacement on trial ... 176*5 tons
I. H. P. on trial 364
Speed 133 knots
Find the "displacement coefficient of speed."
Ans. 203.
4. A steamyacht has a displacement of 143*5 tons, and 250 I.H.P.
is expected on trial. What should the speed in knots be, assuming a
displacement coefficient of speed of 196 ?
Ans. I2'2 knots.
Horsepower^ Effective and Indicated, etc. 345
5. The Warrior developed 5297 indicated horsepower, with a speed
of 14*08 knots on a displacement of 9231 tons. Find the displacement
coefficient of speed.
Ans. 233.
6. In a set of progressive speed trials, very different values of the
' ' displacement coefficient " are obtained at different speeds. Explain the
reason of this.
7. Suppose we took a torpedoboat destroyer of 250 tons displacement
and 27 knots speed as a model, and designed a vessel of 10,000 tons dis
placement of similar form. At what speed of this vessel could we compare
her resistance with that of the model at 27 knots?
Ans. 50 knots.
8. A ship of 5000 tons displacement has to be driven at 21 knots. A
model of this ship displaces 101 Ibs. At what speed should it be tried ?
Ans. 3 knots.
9. A ship of 5000 tons displacement is driven at a speed of 12 knots.
A ship of 6500 tons of similar form is being designed. At what speed of
the larger ship can we compare its performance with the 5oooton ship ?
Ans. 1253 knots.
10. A vessel 300 feet long is driven at a speed of 15 knots. At what
speed must a similar vessel 350 feet long be driven in order that their
performances may be compared ?
Ans. i6'2 knots.
11. A vessel 300 feet long has a displacement on the measuredmile
trial of 3330 tons, and steams at 14, 18, and 20 knots with 2400, 6000, and
9000 I.H.P. respectively. What would be the I.H.P. required to drive a
vessel of similar type, but of double the displacement, at 20 knots ?
Ans. 13,000 I.H.P. about.
12. A vessel of 3100 tons displacement is 270 feet long, 42 feet beam,
and 17 feet draught. Her I.H.P. at speeds of 6, 9, 12, and 15 knots are
270, 600, 1350, and 3060 respectively. What will be the dimensions of a
similar vessel of 7000 tons displacement, and what I.H.P. will be required
to drive this vessel at 18 knots?
Ans. 354 X 55 X 223 ; about 9600 I.H.P.
13. A vessel of 4470 tons displacement is tried on the measured mile at
progressive speeds, with the following results :
Speed. I.H.P.
847 485
1043 8Sl
1223 J 573
1293 2II 7
A vessel of similar form of 5600 tons displacement is being designed.
Estimate the I.H.P. required for a speed of 13 knots.
Ans. 2300 I.H.P.
14. Verify the figures given for the coefficients of speed of H.M.S. Iris
on p. 319.
15. A vessel of 7000 tons requires 10,000 I.H.P. to drive her 20 knots,
and the I.H.P. at that speed is varying as the fourth power of the speed.
Find approximately the I.H.P. necessary to drive a similar vessel of 10,000
tons at a speed of 2iJ knots.
Ans. 16,000 I.H.P.
346
Theoretical Naval Architecture.
16. Dr. Kirk has given the following rule for finding the indicated
horsepower of a vessel :
In ordinary cases, where steamers are formed to suit the speed, the
I.H.P. per 100 square feet of wetted surface may be found by assuming
that, at a speed of 10 knots, 5 I.H.P. is required, and that the I.H.P.
varies as the cube of the speed.
Show that this can be obtained on the following assumptions :
/V\ 2
(i.) The resistance can be expressed by the formula R = f. S. ( ? \
where /'= 0*265.
(ii.) The propulsive coefficient assumed to be about 45 per cent.
17. Prove that the Admiralty displacement coefficient of speed is the
same for two similar vessels at corresponding speeds, supposing that the
efficiency of propulsion is the same. What other assumption is made ?
1 8. Draw a curve on base of speed of the Admiralty displacement
coefficient of speed for H.M.S. Drake Qi 14,100 tons, whose curve of I.H.P.,
based on the trial results, give the following figures :
Speed in knots
10
12
14
16
18
20
22
24
I.H.P.
195
3200
4800
7000
10,000
I 4 ,800
21,900
31,000
\vt x v
299
315
334
34i
340
315
284
260
I.H.P.
Find also the values of the index n in the formula
I.H.P.=*.V (,, = log I.H.P. log I.H.P.,N
log V,  log V, J
Ans. (n k) 271 (13 k) 263 (15 k) 283 (17 k) 303 (19 k) 373
(21 )4'ii (23 ) 40.
19. Using the W* coefficient of speed, determine the I.H.P. of a
vessel similar to the Drake, 555 feet long, at 25 knots, the Drake being
500 feet long. Ans. 42,600 I.H.P.
(For further examples, see 28, 30, 31, and 32 in Appendix A.)
20. A model 20 feet long, wetted surface 77 square feet, has a resistance
of 26*2 Ibs. in fresh water at a speed of 5 knots. Calculate the effective
horsepower in sea water of a ship having 16 times the linear dimensions
of the model and 20 knots speed (/ for model = 0*0104 and for ship
0*0089, speeds in knots).
(Durham B.Sc. 1910.)
The frictional resistance of the model is
0*0104 x 77 * 5 183 = I 5' 2 lbs 
The frictional resistance of the ship at 20 knots is
0*0089 X 77 X i6 2 x 2o 183 = 42,100 Ibs.
The residuary resistance of the model at 5 knots is 26*2 15*2 =
ii Ibs., and that of the ship at the corresponding speed of 5/^/16, or 20
knots, is ii X i6 3 X 1*025 by the law of comparison, allowing lor the
density of salt water, or 46,200 Ibs. nearly.
Horsepower, Effective and Indicated, etc. 347
The total resistance, therefore, of the ship at 20 knots is 42,100
+ 46,200 = 88,300 Ibs.
The E.H.P. is therefore
88,300x^^5^=5417.
21. A vessel on successive runs on the measured mile obtains the
following speeds, viz. :
27592, 28841, 27965, 28943, 27777, 28426
knots respectively.
Obtain : (i) Ordinary average speed,
(ii) Mean of means of 6 runs,
(iii) ,, of first 4 runs,
(iv) ,, ,, of second 4 runs,
(v) ,, ,, of last 4 runs.
AHS. (i) 28257 ; (ii) 2838 ; (iii) 2837 ; (iv) 28418 ; (v) 2832.
CHAPTER IX.
THE ROLLING OF SHIPS.
NOTE. Throughout this chapter, when an angle is called
$ or < it is measured in degrees ; when it is called or it is
measured in units of circular measure, so that
7T
[So
In dealing with the subject of the rolling of ships, it is neces
sary to consider first rolling in still water. Although a ship
will not under ordinary circumstances roll in still water, yet it
is necessary to study this part of the subject before dealing
with the more difficult case of the rolling of ships among waves.
Unresisted Rolling in Still Water. This is a purely
theoretical consideration because, even if a ship is rolled in
still water, the rolling will sooner or later cease because of the
resistances which are set up and which drain the ship of her
energy. This energy is potential (i.e. due to position) at the
extremity of each roll, and kinetic (i.e. due to motion) at the
middle of each roll. At intermediate positions the energy of
the rolling ship is both potential and kinetic. Work has had
to be done in the first place to get the ship over, and the ship
has then stored up in her a definite amount of potential energy.
This energy is gradually dissipated by the various resistances
which came into operation until finally the ship comes to rest.
In a ship rolling we cannot fix upon any definite axis about
which the oscillation takes place. It appears, however, that
the centre of oscillation or quiescent point is not far from the
The Rolling of Ships. 349
C.G. of the ship, and this point is usually taken as the centre
of the oscillation.
The period of oscillation of a ship from side to side rolling
unresistedly in still water through small angles is given by
Where m is the metacentric height in feet.
g is the acceleration due to gravity in footsecond units,
viz. 322.
k is the transverse radius of gyration of the ship in feet,
denned as follows :
(The moment of inertia of a body about any axis is
found by adding together the product of each
weight and the square of its distance from the axis.
If for a ship this axis is through the C.G., W is the
weight and I the moment of inertia, then k is such
a quantity that I = W X / 2 and k is the radius
of gyration. Expressed in mathematical form
I = W x & = S(w X f).)
The following is the reasoning leading to the above expres
sion for the period, which may be omitted by students not
having a knowledge of the calculus.
The equation of motion of the rolling ship is
where GZ is the righting lever in feet.
/mass moment N / angular \ _ /couple causing^
Y of inertia J x ^acceleration^ ~ ^ the motion J
5? + S GZ =
For ordinary ships the curve of stability for small angles is nearly
straight, and we can say GZ = m . 0, so that
d 2 e m . g
7? + ^ e = ....... < 2 >
This is a differential equation, of which the general solution is
(3)
350 Theoretical Naval Architecture.
dQ d^Q
If 6, , are all the same after time /, then
This is the double oscillation. The period of the single oscillation is
/ 2
therefore given by T = ir . / as stated above.
Equation (3) may now be written
6 = A . sin  . / + B.cos . /
and if the initial conditions are assumed, such that when t = o, the ship is
upright, i.e. 6 = O, and the maximum roll of the ship is 0, this equation
becomes
= & sin ^ . / (4)
or the angle of heel on a time base is a curve of sines.
The expression for the period is seen to be independent of
the angle, and it has been found in actual ships that the period
of roll is practically the same for all angles of oscillation when
these angles do not exceed 10 to 15 each side of the vertical.
This is termed isochronous rolling. It is also to be noticed
that to make the period long, i.e. to increase the time of oscilla
tion, it is necessary to
(i) increase the radius of gyration and, or
(ii) decrease the metacentric height.
An application of the above is seen in the current practice of many
merchant vessels. In some trades, voyages have to be undertaken with
little or no cargo, owing to the absence of return freights. It is necessary
for seaworthiness and the proper immersion of the propellers to sink the
vessel by means of water ballast. This ballast has usually been carried in
the doublebottom spaces, leading to a low C.G. of the ship, and a large
metacentric height. The excessive stability causes a short period, and, in
some cases, has not merely rendered the ship uncomfortable, but actually
unsafe. The practice, therefore, has grown up of providing spaces for the
water in other places. Sometimes deep ballast tanks are provided. In one
patent the triangular space at the side beneath the main deck is made into
a ballast compartment, and in another the tank top is continued upwards
to the deck, forming an inner skin at the side, and in the space thus formed
the water can be carried. It is to be observed that such spaces, exclusively
devoted to water ballast, are exempt from the measurement for tonnage.
The added weight of the ballast produces sufficient immersion for seaworthi
ness, but does not give excessive stability, and the weight at the sides tends
to lengthen the period by increasing the radius of gyratiop.
The Rolling of Ships. 351
The following are the periods of some typical ships. It
will be noticed that the heavily armoured battleship of moderate
GM has a long period, about 8 sec. In the deck protected
cruisers with no side armour a quicker motion is experienced,
and in the small classes of war vessel with a relatively large
GM there is very quick motion.
H.M.S. Majestic (about 3^ ft. GM and
great moment of inertia due to side T = 8 sec.
armour)
H.M.S. Arrogant, 2ndclass cruiser, deck) _
protected \
H.M.S. Pelorus, 3rdclass cruiser, deck pro 1 T _ i
tected I
(small period due to (a) small
Gunboats and mQment Qf inertiaj ^ f T = 2 to 4 sec .
Destroyers  latively large GM j
Atlantic liner, small GM T = 10 to 12 sec.
Passenger yachts T = 5 to 6 sec.
/ &
The formula T = TTA /   fails when the metacentric
A/ mm g
height is small. In the particular case of a ship with zero
metacentric height it gives an infinite period for the roll which
is absurd. The problem is possible of solution in a wallsided
vessel, and this was dealt with by Prof. Scribanti at the I.N.A.
for 1904. He took three cases, and for each found T m the
period from his formula and T from the formula above
m.g
V
In a battleship with 3 ft. 1 _T = the errQr being smal]
GM ) TIB
T
In a liner with 4 in. GM . = = 1*31, a considerable error.
In^almost zero GM,
j _T =
352
Theoretical Naval Architecture.
For a ship with zero metacentric height, assuming wall
sidedness, he found by advanced mathematical analysis * the
following expression for the period, viz.
where is the maximum angle in circular measure.
In the case of a vessel having its curve of stability a curve
of sines, like a circular vessel or a submarine, the equation of
motion becomes
jp + 2 . sin = o
This differential equation can be solved by advanced
mathematical methods, and the following are the periods of
single oscillations for various angles from the upright, taking
the period of a small oscillation as unity.
small
30
60
90
120
I'373
I 5
180
I
i 'oi 7
1073
1183
1762
infinite
Thus beyond small angles there is an appreciable lengthen
ing of the period. The range of stability in this case is 180
degrees. If for a ship the curve of stability is of the same
character as a curve of sines, it is reasonable to assume that
when rolling to angles bearing the same ratio to the range as
above a similar lengthening of the period would take place in
comparison with a small oscillation. Thus, for a ship with a
range of 60 the period for a roll of 20 each side of the vertical
would be about 7 per cent, greater than that for small angles.
Although the conditions are not exactly similar to the above
in the case of ordinary ships the period is lengthened some
what for large angles, and this departure from isochronous
rolling has an important bearing on the safety of a ship when
being rolled to large angles in a seaway.
1 Mr. A. W.Johns, R.C.N.C., gave in Engineering, July 18, 1904, a
method of approximate solution of this problem.
The Rolling of Ships. 353
Forces due to Rolling. One application of the equa
tion of motion of a rolling ship is to find the maximum force at
73/1 "2
any point of a ship when rolling. We can say jy f $ = o,
72/3 2
or the angular acceleration = . 6. This is a maximum
when 6 is a maximum, i.e. at the end of a roll. If we take a
ship with a period of 5 sec. rolling through 30, 15 each side
of the vertical, then the angular acceleration at the end of the
2
roll is X 15 X = 01035 in footsecond units. The
25 180
linear acceleration, say 100 feet up, is therefore 1035 ^ n foot
second units. Force = mass x acceleration = X 1035
O
= 032 x weight, i.e. a man 100 feet up would have to hold
on with a force onethird his weight at the end of the roll.
The following is an example of a similar nature worked
out:
A topmast 72 feet in length, height of topmast head being 180 feet above
water, can be assumed of constant diameter, 15 inches. The ship a/8 seconds
period is supposed to roll throtigh 30 each side of the vertical. Make an
estimate of the stress on the material of the topmast at its junction with the
lower mast, supposing it unsupported by stays.
In Fig. 114, w is the weight of the topmast, F the transverse force at
the junction with the mast, L the bending moment, both when at the
maximum at the extremity of the roll. Then we have
(a) resolving the forces at right angles to the mast
w
 F =  (a + h}^
Taking moments about g
( & is for the topmast and = J
from which the bend1
ing moment
2 A
354
Theoretical Naval Architecture.
Taking j 2 = 7. sin 9, we have
w is 3520 Ibs., taking the wood of topmast as 4olbs. per cubic foot, a 36,
h = 108, sin 6 = %
.'. L = 111,000 ft. Ibs. nearly,
from which, using  = p the stress at the base of topmast works out to
4000 Ibs. per sq. in.
FIG. 114.
Resisted Rolling in Still Water. It has been found
by experiment that the rolling of a ship is practically iso
chronous, although resistances to the rolling motion are in
operation. Experimenters on this subject have actually rolled
ships in order to investigate the laws which govern the motion.
A small vessel can easily be set rolling by heaving down with
tackles from a quay. In a large vessel bodies of men can be
run from side to side, their motion being timed to the ship.
In the rolling experiments on H.M.S. Revenge (I.N.A. 1895),
the barbette guns were also trained 15 degrees each side, the
The Rolling of Ships. 355
guns being run out first to make the C.G. eccentric. When
the desired angle of roll was reached, the men and guns were
stationed at the middle line, while the observations were being
taken.
Observing the angles reached on successive rolls a curve
can be constructed as Fig. 115, the abscissae being numbers of
rolls and ordinates the angles reached to port and starboard
successively. Such a curve is termed a a curve of declining
angles. Fig. 116 shows samples of what is termed a curve of
extinction, which is obtained from the curve of declining angles,
the abscissae representing angles of roll and the ordinates
angles lost per swing.
It has been found by analyzing a number of these curves
that the decrement or angle lost per swing can be expressed
as a<j> + ^$ 2 where $ is the angle of roll in degrees and a and b
are coefficients which vary for different ships. Thus, calling
A$ the decrement, we have
A$ = 0$ + $ a
Taking A as a single roll, we have 
I g = * + v
and in the limit this becomes
which is termed the decremental equation.
Thus we have
d&
Inconstant, T = 8 sec. . . .  = 0035$ + 0*0051$*
d<b
Devastation^ T = 675 sec. . ^ = 0072$ f 0*015$*
, wHhout bilge keels, j _^ = o . oi ^ +
wUh^bilge keels, j _g = O . o6 ^ +
The integral form of the decremental equation is
dn =
356
Theoretical Naval Architecture.
5 10 15 2O 25 30 35 4O
WtCrriber of rolls
FIG. 115. CURVES OP DECLINING ANGLES, Revtnge,
K>
I
47,'
f
'^'
*;
IO
15
ANGLE OF ROLL
FIG. 1 16 CURVES OF EXTINCTION. Revenge*
The Rolling of Ships. 357
which gives the number of rolls to pass from an angle of roll fa to an angle
<>!. On integrating this becomes
Thus for a ship in which a = 005, b = 0*02, starting from 15 degrees, 13
rolls are necessary before the angle of roll is 2 degrees. For the Inconstant,
starting from 15 degrees, the successive angles of roll are 13 '5, 12*2, etc.
If a ship rolls from 1 port to 2 starboard, supposing the
curve of stability a straight line, we have
Dynamical stability at i = J . W . m . ?
at 2 = i.W.^. 2 2
so that the loss of energy = JW . m(f  2 2 )
= W . m . m X decrement
taking J(j + 2 ) = m .
If R be the moment of the resistance to rolling at angle B,
,,
the work done by the resistances from x to 2 is / R . dB.
2
We can then equate this work to the loss of energy, since these
must be equal, viz.
R . dB = W . m . m x decrement,
i
and putting in the value of the decrement from the decremental
equation above (and remembering that <1> = J
= W. m.
(
J
This was the method adopted by the late Mr. Froude to
investigate the laws underlying the resisted rolling of ships.
i Suppose the moment of resistance varies as the angular
JJQ
velocity, or R = k\ . . Then assuming, as in unresisted
rolling, B = j . sin ^ . /, we have
dB TT TT
^=i. T COS T ./.
358 Theoretical Naval Architecture.
The work done from to zero is
7/J
on putting in the above value for ^ and integrating. Similarly
for the other side the work done from o to 2 is J . t . ;p. 2 2 .
So that from x to 2 the work done against the resistance is
J . ki ^ . m 2 , putting m 2 = (/ + 2 2 ). Equating this to the
loss of dynamical stability, viz. W . m . TO X decrement, we
have
2
decrement = \ . k^ . ^ rp . OT
jn
i.e. a resistance which has a moment proportional to , the
dt
angular velocity, will give a decrement proportional to the angle of
roll.
2. Suppose the moment of resistance varies as the square
of the angular velocity, or R = k* . \jjj Then, by a similar
process to the above and equating the work done by the
resistance to the loss of dynamical stability, we get that
decrement = f . k, . w ^ ^ z . m 2
i.e. a resistance which has a moment proportional to the square
(dQ\
of the angular velocity \^J will give rise to a decrement pro
portional to the square of the angle of roll.
We thus see that if the resistances to the rolling motion
dQ /d0\
are assumed to vary, partly as r and partly as l~ I , the
decrement is given by
2 ^2
This is of the same form as the decremental equation found
to fit the curves obtained from rolling experiments.
The Rolling of Ships. 359
Mr. Froude attributed the first term to the formation of
waves, and the second to friction and the passage through the
water of bilge keels or keel projections (including the flat
portions of the ship).
Waves. The rolling motion of a ship creates waves on
the surface of the water, and these waves pass away and re
quire energy for their creation. A wave of very small height
represents a considerable amount of energy, and the drain on
the energy of the rolling ship is a distinct resistance tending
to reduce the rolling motion.
Friction. This is of small amount, because the surface
of a ship is kept smoothly painted to reduce the resistance to
steaming to a minimum.
Form of Section. If a ship has a sharp bilge, the water
at the corner has to slip past, and gets a motion opposite to
that of the ship. The effect both as regards friction and on
bilge keels is therefore greater than if the section were more
rounded in form.
Air Resistance. The resistance of the air to rolling is
only small under ordinary circumstances, but it may be made
considerable by the use of steadying sails. If a ship with sails
set rolls to windward, the wind pressure is increased owing to
the greater relative velocity, and this the more so the higher
up we go. The pressure on the sails therefore is greater when
rolling to windward, and the centre of pressure is higher.
When rolling to leeward the effective pressure is less, and there
is a fall in the centre of pressure.
Bilge Keels. Mr. Froude, in his investigations, took the
bilge keels as flat surfaces moving through the water, and by
using data obtained from swinging a flat board in water was
able to make a calculation for the resistance offered by the
bilge keels to the motion (see later for the details of this). It
was found, however, that the observed decrement due to the
b . < 2 portion of the decremental equation could not thus be
accounted for. Professor Bryan, in a paper before the I.N.A.
in 1900, gave further investigations on the subject. Consider
the flow of water round a rightangled bend as Fig. 117. The
water adjacent to the surface has to come to rest at the corner
36o
Theoretical Naval Architecture.
and change the direction of its flow. Thus along AB we get
a diminution of the velocity of the stream lines. With this
decrease of velocity there must be associated a rise of pressure
both along AB and BC. Taking now the case of bilge keels
projecting from the surface as Fig. 118, the ship being sup
posed to be rolling clockwise. The relative velocity of the
ship, and the water along AaAj has to be brought to zero at
A! and there is caused a rise of pressure along A 2 Ai and
similarly along A 4 A 3 . These pressures will have resultants as
FIG. 117.
FIG. 118.
P and Q, which with ordinary shaped sections will give a
moment tending to stop the motion. The effect will be more
pronounced as the section of the ship is sharper, because of
the greater relative velocity of the water past the bilge as
compared with a round section.
Figs. 115, 116 show very clearly the influence of bilge keels
in reducing rolling. It was found in the Revenge^ starting in
each case from 6, that
without bilge keels 45 to 50 rolls were necessary to reduce to 2
with bilge keels 8 2
Curves are also given, showing the effect of motion ahead on
the rolling. In this case the vessel was proceeding into water
that was undisturbed by the rolling motion of the ship, and
the resistance to rolling was somewhat greater than when the
vessel had no onward motion.
It has been found that the addition of bilge keels adds
slightly to the period of rolling, in the case of the Revenge
about 5 per cent.
The Rolling of Ships. 361
The following is the investigation regarding the work done
by a bilge keel or flat surface, on the assumption that the
pressure varies as the square of the speed of the bilge keel
through the water.
Let A be the area of one side of the bilge keels in sq. feet,
r the mean distance of the centre of oscillation,
c the coefficient of normal pressure at i ft. per sec :
in Ibs. per sq. foot.
/ JQ\ 2
Then pressure = c . A . r 2 . ( j } at any instant.
Moment = c . A . r 3 . ( r. }
\at /
(The 2 in the former investigation is therefore c. A . /*.)
7T 2
The decrement is given by t*^b*w~~ ^2 m 2 for a
resistance whose moment is proportional to ( ^ j as seen
above, or
decrement = f . c . A . * . 2 . M 2 (W in Ibs.)
Putting W in tons and T = TT* we have
V m .g
c A . r 3 . P
decrement
which is increased, as one would expect, by increase in the
area of the bilge keels and in the lever. It is also noticed
that the decrement varies inversely as the I of the vessel, so
that the bilge keels are proportionally less effective in a vessel
with large I than with small. The decrement also varies as
the square of the arc of oscillation, so that when large angles
are reached, as in a sea way, the influence of bilge keels will
be most effective.
Boiling among Waves. A wave is not the passage of
water, but the passage of motion. The actual movement of
the particles of water composing a wave is small. The form
moves with considerable speed, but if a piece of wood be
362 Theoretical Naval Architecture.
observed, it is noticed to oscillate about a mean position. In
the generally accepted trochoidal theory the particles of water
for deepsea waves are supposed to move in circular orbits,
and the diameters of these orbits decrease as the depth
increases. This orbital motion gives rise to centrifugal force
and the pressure at the crest of a wave is less than in still
water, and at the trough the pressure is greater. The buoy
ancy, therefore, in the crest portion is less than the normal,
and in the trough portion it is greater. This is the explana
tion of the tenderness of sailingboats on the crest of a wave.
The virtual weight is less than the actual, and so the righting
moment is reduced as compared with still water. The heeling
moment due to the wind is not affected in this way, and so a
boat of sufficient stiffness in still water is liable to be blown
over on a wave. The virtual force of gravity therefore varies
at different places on a wave, and its direction also varies,
being perpendicular to the wave profile at any particular point.
This direction is termed the virtual upright^ and a small raft
will always tend to place its mast along this virtual upright.
This has its maximum inclination to the vertical at about a
quarter the length of the wave from the crest or trough. A
ship rolling amongst waves will at each instant tend to place
her masts parallel to a virtual upright, and a surface which is
normal to each of these virtual upright positions of a ship in a
wave is termed the effective wave slope^ which is distinctly flatter
than the actual observed wave profile.
In dealing with the subject, it is not usual (except for
sailingships) to consider the variation of the amount of the
virtual weight, but allowance must be made for the variation
in its direction. Certain assumptions have to be made to
bring the problem within the scope of mathematical treat
ment. These are as follows :
(a) The ship is lying passively broadside to the wave
advance.
(b) The waves are assumed to be a regular series, identical
in size and speed.
(c) The waves are assumed long in comparison with the
size of the ship.
The Rolling of Ships.
363
(d) The profile of the wave is taken as a curve of sines.
We have first to express the angle the virtual upright makes
to the upright in terms of the time and other known quantities.
Fig. i :i 9 represents the construction of the wave, L being
the  length, H the height (much exaggerated), x and y the
coordinates of a point P referred to axes through the trough.
FIG.
119.
This point P is reached in time /, the time from crest to crest
being 2T 1} i.e. Tj is the half period of the wave.
7T L H H 7T
Then a = ^ . / ; x = ^./; y =  .cos^./
Therefore
dy TT . H
where B l is the slope at P. The slope being small, we may
say that
_ 7T . H 7T
*> =' 117 sra T, '
which is also the inclination of the virtual upright to the
vertical.
From Fig. 120 the equation of motion is, 6 being the angle
of ship from the vertical
W // 2 /9
_^^+W.. (900 =
0_ X being the angle from the virtual upright,
72/1 2
or 2 + (0  0J = o
Theoretical Naval Architecture.
T being the period in still water. Putting in the value of 0,
found above
which is Fronde's general equation for unresisted rolling among
waves.
FIG. 120.
Assuming for the initial conditions, i.e. when / = o that the
ship is upright and at rest in the wave trough, the solution of
this differential equation is
T 7T
i being the maximum wave slope.
i. Take the special case when T = T : , i.e. the stillwater
period of the ship equals the half period of the wave. This
is termed synchronism. Putting T = T! in the above equation
and using the method of the calculus for dealing with inde
terminate forms, we have
,
= 
./cos
1
./J
When
= ,, etc. 6 = ^  y, etc., showing that the
inclination of the ship is alternately half the maximum wave
slope.
The Rolling of Ships.
365
When / = o, T x 2T, etc., we have 6 = o, . 15 TT^
^j, etc., *.*. for every half wave that passes an additional
angle = J . TT x maximum slope, is given to the roll, and thus
a ship under the given assumptions must inevitably capsize (see
Fig. 121). Thus the Devastation with a still water period of
6f", if lying broadside with no resistance, to waves of J
maximum slope and period 13^", would increase the roll every
half wave by . degrees, and in 67 J seconds, or rather over
a minute, would reach 8. Large angles are soon reached
also, if there is only approximate
synchronism between the ship and
the wave. Thus a ship of 5" period
rolling unresistedly broadside to waves
of 4" half period with 8 maximum
slope will, in successive rolls, reach
11, 20j, 2 7 J.
These results are borne out by the
experience of ships at sea. It has
frequently been observed that ships
with a great reputation for steadiness
occasionally roll heavily at sea. This
is due to the fact that a succession of
waves has been met with, having a
period approximately synchronizing
with the double period of the ship.
The synchronism may be destroyed
by altering the course, since what
affects the ship is the apparent period of the waves.
2, Suppose the ship has a very quick period as compared with
T
that of the wave, so that ^r is small. The equation above
then reduces to
FIG.
i.e. the ship takes up the motion of the wave and behaves
366 Theoretical Naval Architecture.
exactly like a raft. The angle of maximum heel will be the
maximum slope of the wave.
3. Now take the case in which the period of the ship is
T
long compared with that of the wave, i.e. f^ is small. The
equation above can be written
rri ~ rr\ *
= j . '[sin ^ . /   1 . sin ^ /J
T
This is always small since ^r is small, and the ship will
never depart far from the vertical. Thus, to secure steadiness
at sea it is necessary to make the stillwater period as long as
possible. To do this there must be a small metacentric
height. Such a ship is crank, i.e. easily inclined by external
forces, but in a sea way is most likely to be steady.
Atlantic storm waves are about 500 to 600 feet in length,
and have a period of 10 or n seconds (i.e. 2 . TJ. The
longest wave recorded had a length of about 2600 feet, and
a period of 23 seconds. The battleship and liner, quoted
above as having periods of 8 and 1012 seconds respectively,
should therefore prove steady ships in a sea way, as synchro
nism would only be experienced when meeting with waves of
periods 1 6 to 24 seconds, which are quite exceptional.
Resisted Rolling among Waves. If we take the
critical case of a vessel meeting with waves whose half period
is equal to the ordinary period of the ship, then the angle for
7T
each swing is increased by  . $ a as seen above. The decre
ment due to resistance is given by 0< + b<$ and the increment
per swing is therefore . $1 (a*j> + & 2 ). The angle of swing
will go on increasing until an angle of roll is reached such
7T
that  . 3>j = a& + ^ 2 . The increase due to synchronism is
then just balanced by the decrement due to resistance, and we
get a steady roll of 3>. We have seen above that when large
angles are reached a ship is not isochronous in her rolling,
and also that the fitting of bilge keels causes an increase in
The Rolling of Ships. 36;
the period. Therefore, under the actual conditions obtaining,
with a synchronous swell the ship will not necessarily capsize, for
(a) As large angles are reached the ship departs from
isochronous rolling.
(b) Resistances come into operation, and there is also
the further condition, viz. :
(c) A succession of waves of precisely the same period
is a very unlikely occurrence.
Apparent Period of Waves. We have spoken above
about the apparent period of waves as affecting a ship's
rolling. If ft is the angle the direction of the ship's advance
makes with the crest line of the wave, then if v be the speed
. of the ship, v . sin ft is the speed of the ship against the wave
advance \ and z/ being the speed of the wave, the waves will
meet the ship at a speed z> + v . sin ft. If T be the actual
period of wave, then the apparent period T' is given by
T = TO 4 (i + ^ sin ft). If ft is negative, ft' say, i.e. the ship
travels away from the wave advance, T' = T f (i . sin/3').
Thus in the first case the apparent period is diminished and in
the second case increased.
Graphic Integration of the Rolling Equation.
i. Unresisted Rolling in Still Water. The equation of motion
of a ship rolling unresistedly in still water has been seen to be
^0 g
^ + ^GZ = o
This cannot be mathematically integrated, because there is
no relation between GZ, the righting arm and the time. By
assuming that GZ = m . 6, a. solution can be found leading to
/ A3
the expression, T = TTA / , being small. This enables
V m.g
the equation of motion to be written
^0,^ GZ_
<// 2 + T 2 ' m =
where T is the time of oscillation from side to side, or one
half the mathematical period.
368
Theoretical Naval Architecture.
By assuming GZ = m . sin 6 we have the case of a circular
ship or a submarine, and the equation can be solved by
advanced mathematical methods, the solution for various angles
being given on p. 352.
By the process known as " graphic integration " the solution
can be accurately found and the process can be extended to
the case when resistances operate.
FIG. 122.
To lead up to the subject, take the case of a body falling
freely under gravity. The force causing the motion is constant,
viz. that due to gravity = P, say. Now force = mass x accele
ration, or P = /i and/= ^, /. v f P . dt taking unit mass.
b
dt
If therefore we have a force curve on base of time, Fig. 122
(in this case a straight line), the velocity is found by integrating
the force curve. Again v = r. or s = jv dt, i.e. the space or
position is found by integrating the velocity curve. Thus, in
the figure the body would have fallen the distance 4D in time
The Rolling of Ships. 369
*, = 4 seconds. EH being the force curve, OBK the velocity
curve, and OBD the position curve.
Now take the reverse process. Having given the position
curve OD, at any point, as D, the tangent makes an angle with
the base 6 such that tan 9 = ^ , i.e. the velocity. At times
o, i, 2, 3, 4, etc., seconds, the velocities are o, g, 2g, 3^, 4^, etc.,
i.e. tan 6 has values o, *, 2g, etc. Setting down as in lower
figure, the tangents to the position curve at A, B, C, D, etc.,
are parallel to Oa, O, O, O</, etc.
The position curve is the second integral from the force
curve, and conversely the force curve is the second derived
from the position curve, and the intersection of tangents at two
points of the position curve is below the centre of gravity of the
corresponding portion of its second derived ', i.e. the force curve.
(For proof of this see later.)
Thus to get A we find the C.G. of OE, and at T draw TAV
parallel to Oa in the lower figure. For the second interval we
draw OAW parallel to O^, and so on with succeeding intervals,
which enables the position curve OABCD to be drawn in.
The equation we have to solve is
J^? = 7r! GZ
df ~T 2 ' m
dropping the sign,
dm TT GZ
^ = T~ 2 '^r
dO .
to = , the angular velocity in circular measure.
In a small time A/ the change of angular velocity is therefore
"* GZ A,
Aw = j A/
T^ m
It is more convenient to use degrees, so that
7T 2 180 GZ
Aw = A/ degrees,
1 TT m
and taking the interval of time . T,
1 80 GZ
2 B
370
Theoretical Naval Architecture.
If therefore at a certain interval the angular velocity is
represented by the slope of the line OB = tan a, Fig. 123, on a
base AO = i'oi3 T, and BD is set up equal to the mean
value of over the succeeding interval ^j T, then the
FIG. 123.
FIG. 124.
slope of OD given by tan /? is the angular velocity at the
end of the interval ^ T, for
/i8o GZ\
AK H &IJ . \ TT ' tfl
tan
AO
BD V
= tan a +
m /
IOI3 1
If the force curve in this case is one where ordinates are
' and if force and position curves are as shown in
TT m
Fig. 124 on a time base, then at any ordinate at time / the
180 GZ
position must correspond to the value of at that
angle. Where these two curves intersect on the base line, as
they must do simultaneously, we have the value of the half
time of the oscillation supposing we start from : the initial
angle. At the angle OA there is a definite value for ^ 
= OB, and we now by a process of trial and error have to
find the curves AP and BP. In the first place we draw a
The Rolling of Ships.
371
modified curve of stability on angle base whose ordinates are
values of (the slope of which is 45 to the base line), as
Fig. 125. A convenient scale to use is found to be J" = i,
J" = i unit of force, and i" = ^ T.
Set off equal intervals of time ^ T on the base line, i, 2, etc.
2O 4O 6O 8O IOO
MODIFIED STABILITY
CURVE
FIG. 125.
FIG. i5A. Graphic integration for a simple pendulum.
(Fig. 1 25 A), and mark off OA equal to the initial angle assumed.
Then at angle O A the modified curve of stability has the value
T
which is set down OB. Now over the first interval the
10
force will vary (diminishing if we start before the angle of
maximum stability and increasing if we start after this angle).
372 Theoretical Naval Architecture.
An assumed slope BD is taken for the force curve. The mean
value over the interval is LK ; the slope given by LK i 1013 T
T
will give the change of angular velocity in interval . Draw
AC parallel to base line and the position curve must be a
tangent to AC at A, since velocity at starting from the extreme
angle is zero. The e.g. of the force curve OD is then found
and squared up to meet AC in C. From C draw CGF parallel
LK
to the slope given by tan a = . Now we check to see
1*0131
if at the angle iG the ordinate of the modified curve of stability
is iD. If not, the process must be repeated until an agree
ment is found. We then proceed to the second interval and
guess in DE and find F over the e.g. of lE. FH is drawn
MN
parallel to the slope given by tan fi = tan a f ^ 2H and
2E are again checked as before. In this way, by a process of
trial and error, the position curve and the force curve may be
obtained and faired in. They must meet simultaneously on
the base line.
Fig. 1 25 A shows the diagram worked out for the case of a
simple pendulum, or a submarine or circular vessel, for which
the equation of motion is
#0 , 7T 2
^a + ,p sm = o
and the ordinate of the modified curve of stability is sin 6.
TT
The initial angle OA is taken as 120, and it is seen that the
curves cross the base line together at an abscissa of o'685T and
double this, viz. i*37T, is the period of the single swing from
120, T being the time of a small oscillation. The working
of this problem for various initial angles is recommended as
an interesting exercise, the results for 30, 60, 90, 120,
150 should be I'oiy, 1*073, I ' I ^3) J '373i 1762 times T
respectively.
Example. H.M.S. Devastation, with GM of 4 feet, has a curve of
stability whose ordinates every 5 are o, 036, 0*69, O'8o, 0*82, 078, O'66,
The Rolling of Ships.
373
0*44, O'2O, and range 43?. The period T for small angles is 675 sec.
Find the period if heeled to (a) 20, (l>) 30 from the upright and allowed
to roll freely.
Ans. (a) 81 sec., (b) io'3 sec.
In the above we have to find graphically the C.G. of a
trapezoid with reference to an ordinate. This is found as
follows : Make AC = J . AE (Fig. 126), DQ being the middle
ordinate. Join CQ and draw SH parallel to the base line,
then H is in the same abscissa as the C.G. of the trapezoid.
Proof. The shift of G. from middle ordinate is due to the shift of the
triangle QS to the position aRQ through a distance of if . h, 2h being the
base. The area of triangle QS is . h . PQ. The moment of transference
is therefore 3 . PQ . A 2 . This equals (area of rectangle) X (shift of C.G. x).
J
or
h 

This may also be proved from the result of Example 22, Chap. II., which
is left as an exercise.
FIG. 126.
FIG. 127.
Proof that the intersection of tangents at two points of a curve
is at the abscissa of the centre of gravity of the corresponding
portion of its second derived curve. Let the curves EFG, OCD,
OAB (Fig. 127) be three curves such that any ordinate of OCD
is equal to the integral of EFG up to that point, and any
ordinate of OAB is equal to the integral of OCD up to that
374 Theoretical Naval Architecture.
point. Then the tangents to the curve OAB at the points A and
B will intersect at the abscissa of the centre of gravity of the
area HG.
EFG is the second derived from OAB.
Let equation of EFG bejy =/(#)
OCD be ;>=/'(*)
OAB be y =f'(x)
Then we have /' (*) = jf(x)dx, and/"(*) = Jf(x)dx
or/w = (x} and/ ' (x) = f " (x)
Take two abscissae x 1 and x 2 for which the ordinates of EFG
nd^ 2> of OCD are ^/ and j' 2 ' and of OABj^/' and y* .
The abscissa x of the C.G. of HG is given by
__
r(f\.
J *i\dx}
Now the tangent to OAB at A has the equation
> *(&?**
and the tangent at B has the equation
The Rolling of Ships. 375
Solving for x will give their intersection, or
). <).+*
dx/ z >!/*/,
which is the same expression as the abscissa of the centre of
gravity of HG found above.
We now have to consider the case of : 2. Resisted Rolling in
Still Water. We have seen that with resisted rolling the decre
ment on a single swing can be written
in circular
2 <?.
or ~ ' = 4 
i.e. a and in the decremental equation d$ = a$ f ^^ 2 can
be written
, __ 4 7T
" 5 ' 2 '
W . m . T 2
from which ^ and k z can be determined if a and b are known.
7 n //7fl\%
The moment of resistance = & . + / 2 . ( )
a / \ a / /
Substituting for the unknowns ki and / 2 we have
Moment of resistance
n circular
measure
. T d& . , T 2 , /^d> \ 2 1 d<b C in de
The equation of motion of the rolling ship is now
W 2 d*Q
g ' ' dt z
+ W . GZ = o
376 Theoretical Naval Architecture.
d z e 7T 2
and for in degrees we have
= o
J
In the former investigation we multiply GZ by , and we
do the same for the resistance, which becomes
2 T d<j> T 2
~^' a '~dt + *v j
which is termed the "resistance indicator" This has to be
added to the modified force, if the ship is swinging away from
the upright when resistance acts with stability in stopping the
motion, and subtracted if the ship is swinging towards the
upright when resistance acts against the stability.
If ? *= 10 degrees per second, then the ordinate of the
eft
resistance indicator is
2 T 5 T 2 .
__,a. IO + f ._.. IOO
This would be set off as an ordinate at a point C (Fig. 128),
such that AC J 1*013 T = 10 degrees per secondhand so on.
Guess in CF resisted and CG unresisted. AD is the mean
force over the interval, and we set up OL = AD. Then the
slope of LQ gives the angular velocity at the end of interval,
MP is therefore the slope of position curve. Then the angle
BN should give on the modified stability curve the distance
BG, FG being equal to the ordinate of resistance indicator at
L. Thus by a process of trial and error we obtain a series
of tangents to the position curve and this crosses the base line
simultaneously with the stability curve. Proceeding past the
upright we should obtain not only the time of the single
oscillation but also the angle from the upright to which the
vessel rolls. The values of a and b for the resistance indicator
are obtained from the decremental equation for the ship.
The Rolling of Ships. 377
As exercises, the " Inconstant " may be taken whose decre
mental equation gives a= 0035, b = 0*0051. Starting from 30
FIG. 128.
IO/3J
the angles reached in successive rolls are 252, 2 1 6. Starting
from 25 the angles reached are 213, 187. T = 8 sec.,
378
Theoretical Naval Architecture.
GM = 2*3 feet. Stability curve at 10 intervals o, 0*5, 1*03,
i*7> 2 '43. 2 '75> 2 ' 6l > 2 * 6 > i'55
3. Unrests ted Rolling among Waves. For a ship broadside
on to a given wave the stability at any instant is determined
by the angle between the centre line of the ship and the virtual
upright, i.e. the normal to the wave surface. This is the angle
between Oa and Ob in Fig. 129. We therefore draw on our
FIG. 129.
base line a curve of wave slope which is taken as a curve of
sines
Thus for a wave of maximum slope 8 and 8" half period we
This then is the base line from
. 180
have i = 8 X sin 5
o
which to measure the angle of inclination to the virtual upright.
The process is then carried on as before. In Fig. 129 is
worked out the case of a ship upright and at rest in the wave
trough with straight line stability on a wave of 80 maximum
slope, and whose half period T x is the same as the time of
oscillation T of the ship. These conditions are known from
previous investigations to lead to an increment of roll to every
half wave of  X 8 = 12^. The tangents to the position curve
The Rolling of Ships. 379
are parallel to the lines drawn from i, 2, 3, etc., to the point
given by i'oi3T. The force curve is checked interval by
interval with the angle from the virtual upright, and must cross
the base line at the abscissa of the intersection of the wave
slope curve with position curve. If the example in Fig. 129
be continued to 20 intervals or over the complete wave, an
angle of 25 would be reached. This is recommended as an
exercise. The process can, of course, be applied for a given
curve of stability and any assumed conditions for period of
ship, period of wave, and maximum slope of wave.
4. Resisted Rolling among Waves. In this case the process
is similar, only the effect of the " resistance indicator " is
brought in as in 2.
For an exhaustive account of the application of the process
of graphic integration, see Sir W. H. White's paper (I.N.A.
1 88 1 ) on the rolling of sailing ships. In this paper, in addition
to resisted rolling among waves, account was taken of (a)
moment due to pressure of the wind on the sails, and (b) the
variation of the virtual weight in different portions of the wave,
this being necessary as this variation affects the righting
moment, while (a) is not thus affected.
Pitching. The expression for the period of pitching of
a ship is of a similar form to that for rolling, but we have to
use /&! the radius of gyration of the vessel about a transverse
axis through the centre of gravity of the vessel, and GM t the
longitudinal metacentric height. This period for a single
oscillation is therefore
It would be desirable, if other conditions allowed, to make
the period of pitching as small as possible, and ships with the
heavy weights concentrated near midships are found to be
better sea boats than vessels with heavy weights at the ends.
EXAMPLES.
I. A vessel of 13,500 tons displacement has a GM of 3^ feet and a
period of 8 seconds. Find the period of roll when 600 tons of coal are
added each side of the vessel in a bunker 21 feet deep and 9 feet wide,
the C.G. of the bunkers being II feet below the original C.G. of the
380 Theoretical Naval Architecture.
ship, and 26 feet out from the middle line. The vessel has a horizontal
curve of metacentres over the limits of draughts corresponding to the above
conditions.
V&
from which & = 823, so that
m
I = 13,500 x 823 = 11,100,000.
The addition of the coal n'down pulls down the C.G. of the ship
I 2OO X II
'  = o'9 ft., making the GM 4*4 ft.
We now have to calculate the new I about the new C.G., I of coal
about old C.G. is given by
2[( T ' 5 X 600 X 9 2 + 600 X 20 7 ) + ( T ^j X 600 X 21* + 600 x ~n)] = 1,008,600
I of total about old G.C. is accordingly
11,100,000+ 1,008,600= 12,108,600
and about the new C.G.
12,108,600 (14,700 x o'9 2 ) = 12,096,700
The new k 1 is therefore
^96,700
14,700
and the new period is accordingly
2. A cruiser of 5000 tons has a metacentric height of 2'8 feet, a period
of 7 seconds, and a horizontal curve of metacentres. Calculate the period
when two fighting tops of 10 ions each are added to the ship at a height
of 70 feet above the C.G.
Ans. 7 '5 sees.
CHAPTER X.
THE TURNING OF SHIPS.
WHEN a ship is moving ahead and the rudder is placed
obliquely to the middle line, the streams of water which flow
aft relative to the ship are deflected in their course and
give rise to a resultant pressure normal to the plane of the
rudder, as P in Fig. 130. The calculation of the amount of
this normal pressure will be dealt with later, but it may be
stated here that it depends on
(i) The area of the rudder,
(ii) The shape of the rudder.
(iii) The angle at which the rudder is placed to the centre
line.
(iv) The square of the speed of the water past the rudder.
The area of the rudder is usually expressed in terms of
the area of the longitudinal middle line plane of the ship, or
approximately the length times the mean draught.
or area of rudder = L.D.
m
The value of m varies considerably. In large war vessels
it is from 40 to 50, but in exceptional cases, where great
manoeuvring power was desired, it came out to 33. In the
Lusitania its value was about 60.
As regards the shape, the pressure for a given area will be
appreciably greater for a narrow, deep rudder than for a broad,
shallow rudder.
The usual maximum angle to which rudders are put is 35
to the centre line.
In a sailing vessel the speed of the water past the rudder
is rather less than the speed of the ship, because there is the
frictional wake. The friction of the water on the surface of
the vessel induces a current of water in the direction of the
382 Theoretical Naval Architecture.
ship's motion so that at the stern the water a short distance
away from the ship has a forward motion. In a sailing ship,
in order to get pressure on the rudder, it is necessary that the
ship shall be in motion, and such a ship loses her power of
steering as she loses way.
In a ship driven by a propeller, although there is the same
frictional wake, the action of the propeller sends a stream of
water astern, so that such a ship has steerage directly the
engines are working, a very great advantage. And this effect
will be greater for a ship having the propeller in line with the
rudder, as in a singlescrew ship and in ships with double
rudders like the Dreadnought^ than in ordinary twinscrew
vessels. Although in screw vessels there is the frictional wake
mentioned above, the speed of the water past the rudder will
be appreciably greater than the speed of the ship, because the
speed at which the water leaves the propeller is greater than
the speed of the ship, the difference being known as the slip.
In any case it is absolutely necessary for good steering
that the water shall get a clean run past the rudder. Vessels
with very full sterns have been found to steer very badly.
In a ship having a deadwood in front of the rudder the
slackening of the speed of the streams of water gives rise to a
side pressure which has a considerable influence in pushing the
ship over at the start. This is specially noticeable in boats.
But for good turning the deadwood is unfavourable, as will be
seen later, and in ships designed for great manoeuvring power
the deadwood is always cut away. See sterns on pages
390, 39i
In Fig. 130 let P be the normal pressure acting on the
rudder at C. Introduce at G, the centre of gravity of the ship,
two equal and opposite forces of value P parallel to the line
of action of P.
Then we have acting on the ship
(i) A couple tending to give angular motion of amount
P X DG ; and .
(ii) a force P acting in the line EG.
The couple is approximately equal to P X X cos 6, and
The Turning of Ships.
383
if P is taken to vary as sin 0, the couple will vary as
sin 6 X cos or sin 26, and it can be shown that this is a
maximum when 6 = 45. The sine law, however, for the
pressure is known to be incorrect, arid it is very probable that
the usual practice of 35 as the maximum helm angle gives
the maximum turning effect. Indeed, it appears quite possible
for certain shapes of rudders that angles of helm as low as 25
will give as good turning results as 35.
The force P acting in the line EG may be resolved into its
components.
(i) FG, tending to move the ship bodily sideways. The
motion in this direction, however, is small because of
the great resistanceof the ship to the side motion.
FIG. 130.
(ii) EF, in a fore and aft direction, and this has a sensible
effect in checking the speed of the ship.
Path of Ship when Turning. On putting the rudder
over the ship will commence to turn in a spiral path, of which
several examples are given later, and this path soon becomes
approximately circular. The distance from the position when
the helm is put over to the position when the ship is at right
angles to her original course is termed the advance. The
distance from the position when the helm is put over to the
position when she has turned through 180 or 16 points is
termed the tactical diameter.
In a ship thus turning, the middle line of the ship points
inside the circular arc and the thrust of the propellers is
accordingly in a direction oblique to the path of the ship.
The resistance to motion is therefore much greater than when
3^4
Theoretical Naval Architecture.
on the straight, and this, together with the fore and aft com
ponent due to the rudder, EF in Fig. 130, causes a very
considerable reduction of speed. In one case the reduction
amounted to quite 50 per cent.
If in Fig. 131 the ship is turning in the path GiGG 2 passing
through the C.G., AF being the centre line of the ship and O the
centre of the arc G 1 GG 2 , then the angle between the tangent
GT and the centre line is termed the drift angle at the point G
If OP is drawn perpendicular to the centre line there is no
drift angle at P, and to an observer on board at P all points of
the ship abaft P will appear to be moving to port, and all
points forward of P will appear to be moving to starboard.
Such a point P is termed a pivoting point, as the ship appears to
pivot about P.
The features of a ship which influence the turning are
principally as follows :
(a) Time taken to put the helm over to the maximum angle.
() Pressure on the rudder.
(f) Moment of resistance of the underwater body of the
ship to turning.
The Turning of Ships. 385
(d) Moment of inertia of the vessel about a vertical axis
through the centre of gravity, to which has to be
added the mass of water associated with the ship.
(a) The introduction of steam steering gear has rendered
this item of less importance than formerly. In ships steered
by manual power the time taken to put the helm over is
considerable, and consequently the possibility of quick
manoeuvring is small.
The general adoption of balanced rudders has facilitated
getting the helm over quickly, as the centre of pressure of the
rudder is close to the axis and the moment required to be
overcome is comparatively small.
(b) The pressure on the rudder depends on various factors,
which have already been dealt with above.
(c) The ship when turning has angular velocity round the
pivoting point P. If we take any portion of the ship at a
distance /from P and of area A, the velocity through the water
j/3
due to the angular motion is I 71 and the resistance varies
/ .jf\\ o
as A f*  lj , and the moment of this about P varies as
A /* ( ft J . This therefore varies as the cube of the length
and the square of the angular velocity. This moment at the
early stages is small and less than the couple caused by the
pressure on the rudder, and consequently the angular velocity
increases. A point, however, is reached when the couple
due to the pressure on the rudder is equal to the moment
of resistance, and then the ship has a constant angular
motion.
It is seen by the above that if areas of the ship under
water can be omitted where / is greatest, the resistance to
the angular movement may be considerably reduced. This
is done by the omission of the deadwood^ or the flat vertical
portion of the ship aft, as the pivoting point is usually well
forward.
The above may be illustrated by the three turning circles
2 C
386
Theoretical Naval Architecture.
given in Fig. 132 of Orlando, Astrcea, and Arrogant. The
profiles of the ships are given, from which it will be seen
that
(i) Orlando has a square type of rudder, not balanced, and
the ship is 300 feet long. The former factor will
delay her entry into the circular path.
(ii) Astrcea has a balanced tudder, and the ship is 320
feet long. The ship gets into the circular path
quicker owing to the balanced rudder, but has a
larger turning circle owing to the greater length.
(iii) Arrogant. Here the rudder area is relatively large,
two rudders being fitted. The length is 320 feet,
as Astray but the stern is cut up considerably.
The influence of these factors is seen in the very
small circle, as compared with the Orlando of smaller
length and the Astraa of the same length.
Many merchant vessels now follow the practice of having
balanced rudders with the deadwood aft cut away. In the
Dreadnought the provision of two rudders with propellers
immediately in front and the cutup shape of the stern (as
Fig. 144) resulted in a marked reduction of the turning circle.
The following is the comparison with two cruisers of nearly
the same length, one having a balanced rudder with no cutup
and the other having a balanced rudder with cutup :
Tactical diameter.
in feet.
Rudder.
Cutup.
In terms
Yds.
of ship's
length.
Powerful
500
Balanced
("None, as\
I Fig. I37J
IIOO
66
Duke of Edinburgh
480
Do.
(As Fig.\
I id. 1 ? I
740
4*63
Dreadnought ...
490
/ 2 No
(Balanced
V L ^J )
As Fig. j
144 /
463
284
The influence on the turning of a ship of a propeller acting
directly on the rudder is strikingly illustrated by the comparative
The Turning of Ships.
FIG.
132.
383
Theoretical Naval Architecture.
turning circles of Topaze and Amethyst. These were cruisers
of similar dimensions, viz. 360' X 40' x 14^' draught, 3000
tons displacement. The
shapes of the sterns
are given in Fig. 133,
from
seen
ship
which it will be
that the former
was a twinscrew
ship and the latter ship
a triple  screw ship.
This latter was the
ship fitted with Par
sons' turbines, with
small screws running
screw immediately in
of the two ships were
FIG. 133.
at high revolutions, and had one
front of the rudder. The rudders
of the same type, viz. balanced, and of much the same
area.
The turning circles are given in Fig. 134, from which it is
seen that the twinscrew ship has a tactical diameter of 870
yards and the triplescrew ship has a tactical diameter of
550 yards, or 7*25 and 46 times the length of ship respec
tively. It is also seen that the latter ship gets into the circular
path much sooner than the former ship. All the condi
tions are practically identical, except that the ship with the
smaller circle has one propeller operating immediately on the
rudder.
(d) The moment of inertia of a ship about a vertical axis
through the C.G. depends on the longitudinal distribution of
the weight, which of course is decided upon for other reasons
than turning. A ship with great weights at the ends will have
a large moment of inertia, and a given turning moment due to
the pressure on the rudder will take longer to get the ship into
the circle than if the weights were more amidships. (It will
be remembered that moment of inertia about any given axis
is found by adding together the products of each portion of
the wnight and the square of its distance from the axis, or
The Turning of Ships.
389
Shapes of Sterns and Rudders. Fig. 135 shows the
ordinary type of rudder fitted to merchant vessels.
In some Atlantic and other liners what is termed the
" cruiser type " of stern and rudder is adopted, analogous
FIG. 134.
to Fig. 143, the principal advantage being the increase
of the length of the waterline obtained for a given length
over all.
Fig. 136 gives the stern and rudder adopted in the
Aquitania. The normal shape of overhanging stern above
390
Theoretical Naval Architecture.
water is obtained, but the deadwood is cut away and a
balanced rudder obtained with the rudderhead below water.
This gives the important advantage of having the rudderhead
and steering gear under water and less liable to damage due
to gunfire. The vessel was built to be an auxiliary cruiser in
case of necessity.
FIG. 135.
FIG. 136.
Figs. 137 to 145 give a number of different shapes of
sterns and rudders adopted in war vessels.
Fig J 37 was adopted for many vessels, including the
protected cruisers Powerful and Terrible. The weight of the
rudder is taken inside the ship, a steadying pintle only being
provided at the bottom of the sternpost.
Fig. 138 was adopted in the Arrogant class, designed as
cruisers to company with the Fleet and in which exceptional
turning facilities were desired. Two rudders are employed, and
the deadwood is cut away.
Fig. 139 is on similar lines with a single rudder.
Fig. 140 is for the Japanese battleship Yas/iima, which had
very great facility for turning.
Fig. 141 was the type of stern adopted for many battle
ships ; the rudder is approximately square, and is unbalanced.
The deadwood is cut away and the sternpost brought down to
take the blocks for docking.
Fig. 142 is the stern adopted in the King Edward VII,
class, the rudder being partially balanced.
Fig. 143 is the stern of the Lord Nelson class, similar to
that of the Yashima.
392
Theoretical Naval Architecture.
Figs. 144 and 145 show the double rudders employed in
most of the Dreadnought battleships and battle cruisers. The
rudder area by this means was made relatively large, and good
powers of turning resulted, in spite of the greatly increased
8A.TTLE CRUISER
FIGS. 143, 144, 145.
length as compared with previous ships. In recent ships two
rudders on the lines of Fig. 138 have been adopted because
of the resistance caused by the bossing out to take the twin
rudders.
Turning Trials. It is usual to carry out systematic
turning trials on H.M. ships, and these are put on record for
the information of the officers.
The Turning of Ships.
393
The following is the method employed to determine the
path of the ship when turning. Two points are selected near
the ends of the ship, at a known distance apart, and at these
positions a horizontal circle graduated in degrees, etc., is set
up with a pointer moveable in a horizontal plane, having sights
which can be kept bearing on any given object as the vessel
swings round. Two weighted rafts with a flag attachment are
dropped overboard about a mile or so apart; it is assumed
that these rafts remain stationary relative to the ship.
The ship is brought up to one of these rafts, at the speed
desired, so as to pass the raft as nearly as can be judged at
L
I
1
TACT i CAL
Di AM E
FIG. 146.
the distance of the radius of the turning circle expected away.
Shortly before coming broadside on, a signal is made, when
the rudder is put over, the course is noted and the time is
394 Theoretical Naval Architecture.
taken, and also the angles shown at the positions forward and
aft, viz. OAB, OB A (Fig. 146), are recorded. These angles
with the known distance AB fully determine the triangle OAB
and consequently the position of the ship relative to the raft.
A similar signal is made at four points (45), eight points (90),
etc., and corresponding observations taken until the ship has
completed the circle. The nine triangles found from this
information are then set out on a convenient scale, as shown
in Fig. 146, and the path of the ship drawn in. The " tactical
diameter " and the " advance " can then be measured off.
Angle of Heel when turning. On first putting a
rudder over, the force on the rudder being usually below the
centre of pressure on the hull on the opposite side, the resultant
couple will have a tendency to heel the ship inwards, but this
tendency is of short duration, as when the ship gets into her
circular path centrifugal action comes into play and an outward
heel results. It is shown in Chap. V. that this heel is
given by
V 2 d
sin 6 = 0088 . 5 qrp
R GM
where V.is speed in knots;
R is radius of turning circle ;
GM is metacentric height ;
d is distance of centre of lateral resistance below the
C.G.
A ship, therefore, of high speed, small turning circle, and
small metacentric height will be liable to heel considerably
when turning at full speed.
Strength of Rudderheads. The formula used by the
British Corporation is as follows :
^=026 4/R . A . S 2
where A is area up to waterline in square feet ;
R is distance of the C.G. of the area from the pintles ;
S is not to be less than IT knots in vessels of and over
250 feet in length.
In vessels of 100 feet, speed taken as 8 knots. Inter
mediate lengths at intermediate speeds in proportion.
The Turning of Ships. 395
Lloyd's Rules do not now give a formula, but give
diameters of rudderheads for speeds varying between 10 and
2 2 knots for different values of A X R as denned above.
Direct Method of determining the Diameter of
Rudderhead. The normal pressure on a rudder of area A
square feet at angle of helm 6 is usually assumed to be
P in Ibs. = ri2 A . ir . sin 6
= 32 A. V 2 .sin0
where v and V is speed of water past the rudder in ft. per sec.
and knots respectively.
It is usual to allow a percentage on to the speed of the
ship to allow for the slip of the screw, although at the stern of
the ship there is the "frictional wake." About 10 per cent,
probably is well on the safe side. V is therefore taken at
ri times speed of ship.
In addition to knowing the pressure, it is necessary to know
the point at which the centre of pressure acts in order to find
the twisting moment about the axis. At 35 the centre of
pressure is taken at threeeighths the breadth from the leading
edge for a rectangular rudder. For other shapes of rudder the
area may be divided approximately into rectangles, or we may
adopt the method given later by dividing into a number of
strips.
Having obtained the twisting moment (preferably in inch
tons), we equate to the formula
T =&../.*
where d is diameter of rudder in inches ;
/ is factor of strength allowed, say
4 tons for wrought iron,
5 ,, cast steel,
3 phosphor bronze.
The following example will illustrate the method :
A rudder is 243 square feet in area, and the centre of pressure is esti
mated to be 6' 1 2 feet abaft the centre of rudderhead at 35. If the speed
of ship is 19 knots, estimate the diameter of the rudderhead if of cast
Steel.
396 Theoretical Naval Architecture,
Pressure in tons =  x 243 X 2O'9 2 x 0*574 = 87 tons
Twisting moment = 87x612x12 = 6389 inch tons
.'. T'S * 5 <** = 6389, taking/= 5
from which d = 187 inches.
Note. If such a rudder is assumed to be a square and supported by two
pintles at the forward edge, one at the bottom and one halfway up, it
can be shown that, where W is the total load and / the total depth, that
Bending moment at head = 3 ! s . W . / 1 , .. f ,
. , , , ITT 7 >both of these are small
middepth = T fo . W./j
Force at head = y W
,, centre pintle = }W
lower pintle = ^ W
The above is an example of where pure twisting only need
be considered (as for Fig. 141), but there are other cases to
consider
" (i) Rudderhead fixed in direction at sternpost, and the
lower part supported at the bottom (as in Figs. 137
and 142).
(ii) Rudderhead fixed in direction at sternpost, and rudder
supported about halfway up, the bottom being free
(as in Figs. 140 and 143).
(iii) Rudder fixed in sternpost and the lower part un
supported (as in Figs. 144 and 145).
In (i) and (ii) both bending and twisting come into play.
In (iii) bending is the determining factor in calculating the
diameter of the rudderhead. It is generally assumed that the
sternpost holds the rudderhead fixed in position. This gives
results well on the safe side.
(1) For the case (i) above, if the rudder is regarded as
a beam uniformly loaded, it may be shown that
' Bending moment at upper end = J x load X depth
Support at head = f X load
heel =  X load.
(2) For the case (ii) above, if the rudder is regarded as a
beam loaded at lower half to twice the intensity of the upper
The Turning of Ships. 397
half (i.e. the rudder is assumed to be a square with the upper
corner cut out), it may be shown that
Bending moment at upper end = ^ X load x depth
Support at head = ~ x load\ in opposite
heel = H x load/ directions.
(3) For the case (iii) above, the bending moment at the
head is found by multiplying the pressure by the distance of the
C.G. below the top.
When both bending and twisting have to be considered,
we equate ^ . TT . f . a* to the equivalent twisting moment, viz.
M + V M 2 + T a , where M is the bending moment and T the
twisting moment.
It may happen that the astern conditions will be the
determining factor, because then the centre of pressure is
nearer the after edge of rudder and farther from the axis than
when the ship is going ahead. The speed astern is usually
taken at half the speed ahead. In any case, in designing a
rudder, sections must be made at various places besides the
p M
rudderhead, and the formula  = j applied to determine the
value of the stress p.
In fixing the shape of a rudder it has to be borne in mind
that at all angles the centre of pressure should be abaft the axis.
For angles below 35 the value threeeighths from the leading
edge does not apply. The following formula may be used,
based on Joessel's experiments for rectangular plates of
breadth b.
C.P. from leading edge = 0^195 b f 0*305 b sin
The formula given above, viz. P = 1*12 . A . iP . sin 0, is
known to be incorrect for small plates moving through water,
and the matter was exhaustively considered by Mr. A. W.
Johns, R.C.N.C., at the I.N.A. for 1904. The formula, how
ever, has been extensively employed for many years for rudder
calculations with satisfactory results. It is to be observed
that
(i) A rudder does not get the full angle at once, so that
393
Theoretical Naval Architecture.
the pressure is not in the nature of a shock. We,
however, use a coefficient of strength giving a large
factor of safety, as if this were the case.
(2) By the time the rudder is over the speed of the ship
suffers an appreciable check.
Centre of Pressure, Calculation of. For rudder
shapes other than rectangular it may be assumed that if the
C.G. is x feet from the midbreadth, the centre of pressure
is x feet from the position it would have if the rudder were rect
angular. Preferably the following method may be employed :
Horizontal ordinates are drawn as shown (Fig. 147),
common interval h, and f the length from forward and after
FIG. 147.
edges is set out on each. Such points represent the centre of
pressure of strips of the rudder at the ordinates. Curves of
centre of pressure are drawn as shown.
Simpson's Rules then are applied, as indicated in the
following table :
The Turning of Ships.
399
Ahead.
Astern.
Products
Ver
Product
No.
Ordi
nate.
S. M.
for
area.
C.P.
from
Product
for
C.P.
from
Product
for
tical
Lever.
for
moment.
axis.
moment.
axis.
moment.
I
/i
I
/i
0,
**
*,
^,
2
/2
4
4/2
^2
4^2/2 >
^2>
4^2 7g>
I
4/2
3
/3
2
2/3
rt s
etc.
etc.
etc.
2
4/3
4
/4
4
4/4
4
3
1 2y 4 ,
5
/5
2
2/5
a s
4
etc.
*
J;
4
2
' ?:
i
8
/8
4
4/8
a
7
9
/9
i
^
8
8,
Area = S l X ^ X A
c;
C.P. astern = from axis
S 2 S 3 S 4
C.P. ahead =  2 from axis
^i
C.P. below top =  4 X h
The size of the rudderhead at the steering gear crosshead
will be determined by the maximum twisting moment ahead or
astern.
The above method may be employed to approximate to
the position of the centre of pressure at smaller angles by the
use of Joessel's formula given above, in order to ensure that
at all angles the centre of pressure is abaft the axis.
CHAPTER XL
LA UNCHING CALCULA TIONS.
BEFORE starting on these calculations it is necessary to
estimate as closely as possible the launching weight of the
ship, and also the position of the centre of gravity both
vertically and longitudinally. The case of the Daphne,
which capsized on the Clyde 1 on being launched, drew
special attention to the necessity of providing sufficient
stability in the launching condition. A ship in the launching
condition has a light draught, great freeboard, and high
position of the C.G. It is possible, by the use of the prin
ciples we have discussed at length, to approximate to the
metacentric height, and if this is not considered sufficient,
the ship should be ballasted to lower the centre of gravity.
It has been suggested that a minimum G.M. of i foot should
be provided in the launching condition. If the crosscurves
of stability of the vessel have been made, it is possible very
quickly to draw in the curve of stability in the launching
condition, and in case of any doubt as to the stability, this
should be done.
It is necessary to prepare a set of launching curves in the
case of large heavy ships, in order to see that there is (a) no
tendency for the ship to " tip," i.e. to pivot about the after end
of the ways (as in Fig. 148 *:), in which case damage would
probably ensue ; and (b) to obtain a value for the force which
comes on the fore poppets when the stern lifts.
In Fig. 148 , if G is the position of the C.G. of the ship,
and B the centre of buoyancy of the immersed portion, then
assuming a height of tide that may safely be expected for
launching
the moment of the weight abaft the after end of ways = W x d
buoyancy = w x d l
1 See Engineering '(1883) for a report on the Daphne, by Sir E. J. Reed.
Launching Calculations.
401
Then for different portions of the travel down the ways the
values of d, w, and d' can be readily found and curves drawn
as in Fig. 149, giving values of (W x d) and (w X d) on a
base of distance travelled. The former will, of course, be a
straight line, starting from the point where the C.G. is over
the aft end of ways. This line should be below the other
curve, and the minimum intercept between them is called the
"Su/
FIG. 148.
" margin against tipping." If it happens that the curves inter
sect, it shows that a tendency to tip exists, and either (a) the
ways should be lengthened, or (b) ballast placed forward, both
of which increase the travel required before G comes over the
end of the ways. The buoyancy curve should be drawn out
for various heights of tide, in order to know the minimum
height of tide on which the ship could be safely launched.
This is a point of importance in some shipyards where tides
do not always rise as high as expected owing to adverse winds.
Ships have been launched successfully which had a tipping
moment, but owing to the speed of launching the danger space
was safely passed ; but this is a risk that few would care to
take.
As the ship goes further down the ways a position is
reached when the moment of the buoyancy about the fore
2 D
402
Theoretical Naval Architecture.
poppet equals the moment of the weight about the fore poppet.
At this point the stern of the ship will begin to lift. If W and w
be the values of the weight and buoyancy respectively at this
point, then the weight W w, instead of being taken over
the length of ways in contact, is concentrated at the fore
poppets. This weight is localized over a short distance both
on the ship and on the slip, and it is desirable to know its
amount and the position on the slip where it will come.
Values of w are obtained at various points of the travel,
and two lines drawn on a base of travel giving values of
2.000.0_Q 10.000
i.ooo.qoo 5.0Q&
300'
W and w, the former being constant. Then if a and b be the
distances of the C.G. and the C.B. from the fore poppet (as
in Fig. 148 a) at any point of the travel
moment of weight about fore poppet = W x a
buoyancy = w X b
Curves are then drawn as in Fig. 149, giving these moments
on base of travel, and the point where these curves cross gives
the position where the stern begins to lift, and the intercept
between the curves of weight and buoyancy at this point,
viz. aa, gives the weight on the fore poppets. In this case
Launching Calculations. 403
the weight on the fore poppets was 2500 tons, the launching
weight being 9600 tons.
The launching curves for H.M.S. Ocean are given in a paper
by Mr. H. R. Champness, read before the Institution of
Mechanical Engineers in 1899. In that case the weight of
the ship was 7110 tons, and the weight on the fore poppets
1320 tons.
The internal shoring of the ship must be specially arranged
for in the neighbourhood of the fore poppets, and the portion
of the slip under them at the time the stern lifts must be made
of sufficient strength to bear the concentrated weight.
Variation of Pressure on the Ways. In addition to
knowing the pressure per square foot when the vessel is on
the slip, it is sometimes desirable to know how this pressure
varies as the ship goes down the ways. It is quite possible
that the pressure might be excessive, and the necessity of
strengthening the slip or shoring the ship internally would
have to be considered.
The support of the ways at any point of the travel is
W w, where W is the weight and w the buoyancy. From
this the mean pressure P can be determined. The support of
the ways must act at a distance x from the fore end, such that
(Ww)x = W.aw.d
where a and b are the distances of the C.G. and C.B. respec
tively from the fore poppet. Let y be the distance of the
centre of pressure from the way ends. There are three cases
to consider (see Fig. 150).
(1) If x lies between \l and /, where / is the length of
surface of ways in contact. Then, knowing P the mean
ordinate, and assuming that the curve of pressure is a straight
line, P A and P F , the pressures at the after and forward ends,
can be determined. (See Example 28, Chapter II.) \ix~\l
or /, then the maximum pressure is 2P, and occurs at the
forward or after end as the case may be.
(2) If x is less than ^/, the distribution of pressure is
assumed a straight line on a base = 3^, and maximum pressure
. 2 W  w
at fore poppet is f . .
404
Theoretical Naval Architecture.
(3) If y is less than \l we have similarly the maximum
W  w
pressure at the after end of way = f . .
By this means a curve of pressure at fore poppet may be
obtained for all positions of the ship on the slip until the stern
FIG. 150.
lifts, P F , and similarly for the after ends of ways, P A , as in
(4) Fig IS 
The maximum pressure at the after end of the ways thus
calculated in one ship was 3^ tons per square foot as compared
with the mean pressure before launching of 1*7 tons.
Launching Calculations. 405
The following papers may be consulted in regard to the
launching of ships :
H. R. Champness, Ocean " I.M.E.", 1899
W. J. Luke, Lusitania " I.N.A.", 1907
J. Smith, "I.N.A.", 1909
A. Hiley, " I.N.A.", 1913
Example. In a certain ship the length of sliding ways was 535 ft.
and the breadth 5 ft. 4 ins. The launching weight was 9600 tons with
C.G. estimated at 2475 ft forward of the after end of sliding ways.
Calculate the mean pressure per square foot on the ways, and assuming the
pressure to vary uniformly as above, calculate the pressure per square foot
at the forward and after ends of sliding ways before launching.
Ans. 1*68 tons j 1*3 tons; 2*0 tons.
(Note. These two latter values are the starting points of the curves in
(4) Fig. 150, the latter rising to a maximum value of 3 tons after a
travel of 250 ft. The former curve rises to a high (indeterminate) value
when the stern lifts.)
APPENDIX A
Proof of Simpson's First Rule. Let the equation of the curve
referred to the axes O.r, Oy, as in Fig. 35, p. 53, be
y = a Q + a\x + a. 2 x 2
a , a^a 2 being constants ; then the area of a narrow strip length y
and breadth &x is
y x A.r
and the area required between x o and x = 2^ is the sum of all
such strips between these limits. Considering the strips as being a
small breadth A.r, we still do not take account of the small triangular
pieces as BDE (see Fig. 12), but on proceeding to the limit, i.e.
making the strips indefinitely narrow, these triangular areas dis
appear, and the expression for the area becomes, using the formula
of the calculus
y . dx
o
or, putting in the value for y given by the equation to the curve
J o
which equals
which has to be evaluated between the limits x = 2^ and x o.
The expression then becomes
Now, suppose the area = AJ/J + By 2 + Cy 3
using the equation to the curve and putting x = o, x = h and
x 2h respectively,
Area = a (A + B + C) + ^(B + 2C) + a 2 . W(B + 4Q . (2)
408 Appendix.
By a wellknown principle of Algebra we can equate the
coefficients of a , a^ and a 2 in (i) and (2), so that
A + B + C = 2h
K.h
B . #
from which A =
so that the area required is
which is Simpson's First Rule.
It may be shown in a similar manner that Simpson's First
Rule will integrate also a curve which is of the third degree, viz.
y = a + a^ . x + a.j? F a s . X s . . . . (3)
Simpson's First Rule is thus seen to integrate correctly curves
both of the second and third degree. It is always used, unless the
conditions are such that its use is not possible.
Proof of Simpson's Second Rule. This may be proved
similarly to the above, assuming that the curve has the equation (3)
above.
Proof of the Fiveeight Minusone Rule. The area between
y l and y 2 is given by
I,
Assuming this to be equal to A^ + B/ 2 + Cy 3 , substituting for
and J 3 , and equating coefficients of , a^ and 2 , we find
so that the area required is
5^.
The area between the ordinates y 2 and y z is
WGto+W''/i)
and adding together, the whole area is
which is Simpson's First Rule.
Proof of the Threeten Minusone Moment Rule (given
on p. 58). Assume the equation of the curve is
y  a 4 ,*
Appendix.
409
Then the moment about \
end ordinate/
n
J o
= yx.dx
= h\\a Q + \ajt + tf 2 ^ 2 )on integrating
Let the moment = (Ajj + By 2 + Cy 3 ).
Substituting for y lt jr zt y 9 the values found from the equation to
the curve, and equating coefficients of a 0) a 1} a. 2) we get
A = ft#, B = !#, C = sVfc 2
so that moment = ^ih\^y l 4 loja ~J 3 )
Simple Area Rule for Six Ordinates (for which neither the
first nor the second rules can be used).
This is obtained by using the Fiveeight Rule for the ends, and
Simpson's Second Rule for the middle portion, thus
I
A, If, If, II, If, A
ff[o'4, i, i, i, i, 04]
Proof of TchebychefF's Rule with Four Ordinates. The
following is the proof in the case where four ordinates are employed.
In solving the equations for eight or ten ordinates imaginary roots
are obtained, but the figures obtained for four and five ordinates
can be combined together for the two halves of the length giving
the figures in the table on page 18.
FIG. 151.
Let the curve BC (Fig. 151) be a portion of a parabola whose
equation referred to the base and the axis OY is
y = a + !.*+ a 2 .x* + a z .x*+ a.x* . . (i)
where a , a lt a v a 3 , and 4 are constants.
(For 4 ordinates the curve is taken of the 4th degree.)
( .1 ., ,. th )
4 IQ Appendix.
Let 2.1 be the length of the area, and select the origin at the
middle of the length.
/"*'
Then the area required = / y doc (2)
J i
r
. . (3)
j j /
Now let this area = C x the sum of the 4 ordinates (4)
= C x
C
.... (5)
< a ....
substituting forj/u^j, etc., their values as given by the equation
to the curve (i), and taking the ordinates symmetrical about OY.
Equating coefficients of a^ a a , a in the equations (3) and (5),
we have C =  and
From these equations we find
jfi = o'i876/
** = 07947/
which gives the positions of the ordinates such that the area
= L(yi + yi + yi + J 2 ), * the summation of the ordinates
4
is multiplied by the length and divided by the number of ordinates.
Displacement Sheet by TchebychefTs Rule. This method
may be extended to rinding the volume of displacement of a ship,
and a table may be employed similar to that on Table I., 1 and
described in Chapter II. There does not appear to be any advan
tage in applying Tchebycheffs rule in a vertical direction, as the
number of waterlines are few in number compared with the
number of ordinates usually employed fore and aft ; and also by
having the waterplanes spaced equally, the displacement and
vertical position of the C.B. for the other waterplanes can be
determined. In the specimen table, therefore, given on Table II., 1
Tchebycheffs rule is employed for the foreandaft integration,
and Simpson's first rule for the vertical integration. The figures
shown in thick type are the lengths of the semiordinates of the
various waterlines spaced from amidships as indicated at the top
of the sheet. These lengths added up give a function of the area
1 To be found at the end of the book.
Appendix. 41 1
of each waterplane, as 2417 for the L.W.P. These functions are
put through Simpson's rule 1 in a vertical direction, and the
addition of these products gives a function of the displacement,
viz. 1 80225. This function, multiplied for both rules, etc., as
shown, gives the displacement in tons, viz. 16,067 tons.
This result is obtained in another way, as in the ordinary
displacement sheet, and an excellent check is thus obtained on
the correctness of the calculation. The semiordinates of the
various sections are put through Simpson's rule, and functions of
the areas of the sections are thus obtained, as 255*05 for the
section numbered II. These functions are then simply added up,
and the same result is obtained as before for the function of the
displacement, viz. 1802*25. It will have been noticed that the
ordinates at equal distances from the midlength are brought
together ; the reason for this will appear as we proceed.
The position of the centre of buoyancy of the main portion
with reference to the L.W.P. is obtained in the ordinary way. To
obtain the position of the centre of buoyancy of the main portion
with reference to the midlength, we proceed as follows. The
functions of areas I. and I A. are subtracted, giving 2*9, and so
on for all the corresponding sections. These differences are
multiplied by the proportion of the halflength at which the
several ordinates are placed, and the addition of the products
gives a function of the moment of the displacement about the
midlength. In this case, the function is 36*4755. This, multi
plied by the halflength and divided by the function of the dis
placement, 1802*25, gives the distance of the centre of buoyancy of
the main portion abaft midships, 6*07 feet.
The lower appendage is treated in the ordinary way, as shown
in the lefthand portion of the table. The reason of this is that
the equidistant sections of the ship are usually drawn in on the
body plan for fairing purposes, and the areas below the lowest
waterline can be readily calculated. The sections at the stations
necessary for Tchebycheff's rule would not be placed on the body
for this calculation, but the ordinates at the various waterlines
would be read straight off the halfbreadth plan.
The summary to obtain the total displacement and position
of the centre of buoyancy is prepared in the ordinary way, and
needs no explanation. The result of this summary is to give
the displacement as 16,900 tons, having the centre of buoyancy
11*2 feet below the L.W.L. and 6*52 feet abaft midships.
1 In this case, instead of i, 4, 2, 4, 2, 4, i, the halves of these are
used, viz. , 2, i, 2, i, 2 , the multiplication by 2 being done at the end.
412 Appendix.
Transverse BM. To determine the moment of inertia of the
L.W.P. about the middle line, we place the ordinates of the L.W.P.
as shown, cube them and add the cubes, the result being 211,999.
This is multiplied as shown, giving 8,479,926 as the moment of
inertia of the main portion of the L.W.P. about the middle line.
Adding for the after appendage, we obtain 8,480,976 as the moment
of inertia of the L.W.P. about the middle line in footunits.
The distance between the centre of buoyancy and the transverse
metacentre is given by^, or
8,480,976
= 1434 feet
16,900 x 35
The transverse metacentre is accordingly 1434 11*2 = 314 feet
above the L.W.L.
Longitudinal BM. The position of the centre of gravity of the
main portion of the L.W.P. is obtained by taking the differences
of corresponding ordinates of the L.W.P. and multiplying these
differences by 00838, etc., as shown. The addition of these pro
ducts, 14*6244, treated as shown, gives 18*15 f et as tne distance of
the centre of gravity of the main portion of the L.W.P. abaft
amidships. Adding in the effect of the after appendage, we find
that the area of the L.W.P. is 29,144 square feet, and the centre of
flotation is 19*53 feet abaft amidships.
To determine the position of the longitudinal metacentre, we
need to find the moment of inertia of the L.W.P. about a trans
verse axis through the centre of flotation. This has to be done in
several steps. First we determine the moment of inertia of the
main portion about amidships. This is done by taking the sum of
corresponding ordinates and multiplying these by (o'o838) 2 , (0*3 127) 2 ,
etc., or 0007, 0098, etc. The addition of the products, 50*5809, is
multiplied by 2 for both sides, by $ for TchebychefFs rule and
by (300)2, being the square of the halflength, because we only
multiplied by the square of the fraction of the halflength the
various ordinates are from amidships, and not by the squares of
the actual distances. The result gives 546,273,720 in footunits
for the moment of inertia of the main portion of the L.W.P. about
the midship ordinate. We add to this the moment of inertia
of the after appendage about the midship ordinate, obtaining
539,399,902 in footunits. This is the moment of inertia of the
L.W.P. about the midship ordinate. To obtain the moment of
inertia of the L.W.P. about a transverse axis through the centre
of flotation, we subtract the product of the area of the L.W.P. and
Appendix. 413
the square of the distance of the centre of flotation abaft amidships.
The result is the moment of inertia of the L.W.P. about a trans
verse axis through the centre of flotation we want, and this divided
by the volume of displacement gives the value of the longitudinal
BM, 927 feet.
The moment to change trim one inch is obtained in the
ordinary way, assuming that the centre of gravity of the ship is
in the L.W.L. and that the draught marks are placed at the
perpendiculars.
To obtain Cross Curves of Stability by means of the
Integrator and using TchebychefF's Rule. The rule we have
been considering can be used with the integrator to determine the
ordinary cross curves of stability in just the same way as with
Simpson's rules. In Chapter V. the process of the calculation
necessary with the integrator is explained. This calculation may
be considerably shortened if Tchebycheff's rule is used instead of
Simpson's rule. Not only can fewer sections be used, but the
integrator itself performs the summation. In this case a body
plan must be prepared, showing the shape of the sections at the
distances from amidships required by the rule. Take, for example,
a vessel 480 feet long, for which by the ordinary method twenty
one sections would be necessary. By using this rule nine sections
will be quite sufficient, and by reference to the table on p. 18 the
sections must be placed the following distances forward and aft
of amidships, viz. 40*3, 126*9, I 44' 2 218*8 feet respectively, the
midship section being one of the nine sections.
The multiplier to convert the area readings of the integrator
employed into tons displacement was for this case 1*097, and to
convert the moment readings into foottons of moment was 13*164.
All that is necessary, then, having set the body plan to the required
angle as in Fig. 79, is to pass round all the nine sections in turn
up to the waterline you are dealing with, and put down the initial
and final readings. We have, for example
Area readings.
Initial 14*198
Final 25,397
11,199 difference
Displacement in tons = 11,199 x 1*097
= 12,285 tons
Appendix,
Moment readings.
.. 5215
Initial
Final
301 difference
Moment = 301 x 13*164 = 3962 foottons
It will at once be seen, on comparison with the example given
on p. 199, that there is a very great saving of work by using this
method. The following table gives the whole of the calculation
necessary to determine a cross curve for the above vessel, values
of GZ being obtained at four draughts, viz. at the L.W.L., one
W.L. above and two W.L.'s below :
Number of W.L.
Area
reading.
Differ
ence.
Displace
ment.*
Moment
reading.
Differ
ence.
GZ
Initial
1,505
__
5,136
_
_
A. W.L.
14,198
12,693
13,924
5,215
79
0*07
L.W.L.
25,397
11,199
12,285
5,516
301
032
2 W.L.
35,126
9,729
10,673
6,373
57
ro6
3 W.L.
43,301
8,175
8,968
7,261
888
130
Displacement Sheet (used in Messrs. John Brown $ CoSs
Drawing Office}. The displacement table used in the drawing
office of Messrs. John Brown & Co., Clydebank, presents several
points of interest, and is admirably designed to conveniently con
tain on one sheet all the calculations necessary for the geometrical
features of a ship's lines. This table was devised by Mr. John
Black, and I am indebted to Mr. W. J. Luke for permission to
reproduce it. The sections are numbered from aft, and the
waterplanes from below (see Tables III. and I HA. at end of
book).
. In Tchebycheff's threeordinate rule (p. 18), the distance of the
ordinates either side of the middle ordinate is 0707 times the half
length of base. If, therefore, we apply this rule five times over for
the length, we should set off on either side of Nos. i, 3, 5, 7, 9
(Fig. 152) a distance of 0*0707 x length, the length being divided
into ten equal parts. The addition of the ordinates A, B, C, . . .
Appendix.
4*5
O, P, Q, multiplied by ^ length, gives the half area of the water
plane.
The ordinates B, E, H, M, P (Fig. 152) are the following
.distances from amidships (6, 3, o, 3, 6) . The ordinates A, C, D,
F, G, K, L, N, O, Q, are the following distances from amidships,
viz. (706, 494, 406, 194, ro6, ro6, 194, 409, 494, 7*06) .
These are so close to the integers 7, 5, etc., that without appreci
able error the integers 7, 6, 5, 4, etc., can be used for the levers as
in the ordinary displacement sheet. Thus, for any waterline the
addition of ordinates in column A multiplied by 2 X will give
the area. The algebraic sum of column B, divided by the addition
of column A and multiplied by , gives the distance of the centre
of gravity of water plane from o to 10 from midships. Column D
is got by multiplying the figures in column B again by the levers,
and the addition of the column properly multiplied leads to the
longitudinal moment of inertia of waterplane about amidships.
This has to be corrected for the after appendage (if any), and then
transferred to the centre of flotation, as explained in Chapter IV.
From this the longitudinal B.M. is readily obtained for the several
waterplanes.
In columns C are placed the cubes of the ordinates in columns
A, and the addition of these columns leads to the transverse
moment of inertia of the waterplanes, from which values of the
transverse B.M. is obtained for the several waterplanes.
The lower appendage is treated by " Thomson's rule," l the
sections used being those on the ordinary body plan. The multi
pliers are obtained as follows :
1 It is understood that ordinary sections and Simpson's multipliers are
now used in this sheet for the appendage.
416
Appendix.
i
1
2
3
4
5
6
7
8
9
9i
10
Area i to 9
Areas o to i, 9 to 10
Whole area
Twice area
Area
I
1
fl
2
I
cm 1 4m M M
2
I
2
I
2
I
i
2
I
2
I
i
{
!
The vertical C.B. of the appendage is obtained by Morrish's
rule, viz. M  +  J, where d is the depth of appendage, v is its
volume, and a is the area of No. I W.P.
In the combination table the results are grouped together to find
the displacement and C.B. up to Nos. 3, 5, 7, and 9 waterplanes.
The displacement is found, in the first place, to the moulded
surface of the ship, as is usual outside the Admiralty service ; the
area of wetted surface is obtained by the formula S = 1 7 L . D +
y
=r, and using a mean thickness of plating, the displacement of the
plating is readily obtained, and thus the "full" displacement is
obtained.
From the results a series of curves as in Fig. 153 is
readily constructed, on base of draught, of displacement, tons per
inch, centres of flotation, transverse metacentre, longitudinal
metacentre, vertical C.B. fore and aft C.B., moment to change
trim one inch, area of midship section and area of wetted surface,
and also the various coefficients. To avoid confusion, the curves
of vertical C.B. and metacentres are measured from the axis marked
L.W.L ; those of C.B. and C.F. abaft amidships are measured from
the right boundary of the figure. All others are measured from the
left boundary. The scales used are appended to all the curves.
These curves, when once carefully drawn for a ship, are of great
value as records of the features of the ship's form.
Loss of Stability due to Grounding. When a ship is being
docked, the shores cannot finally be set up until the keel takes the
blocks all fore and aft. Until this happens there is a portion of the
weight taken by the after block (supposing the ship is trimming by
the stern), and this becomes a maximum immediately before the
ship grounds all fore and aft. Before the shores are set up there
is a considerable upward pressure at the keel which might be
sufficient, under certain circumstances, to cause instability, and
cases are on record in which a ship has fallen over when being dry
docked owing to this cause.
sj.Honvaa
2 E
4i8
Appendix,
In Fig. 154 let the first diagram represent the ship, floating
freely, having a small inclination. In the second diagram a portion
of the weight, w say, is taken by the blocks. This is equal to the
FIG. 154.
displacement between the lines W'L' and W"L". If M, be the
metacentre corresponding to the waterline W"L", then
Moment of stiffness = {(W
 w . OG} sin 6
To find w we can proceed as follows :
1. Accurately. Obtain the displacement and longitudinal
position of C.B. when floating freely. At the instant of taking the
blocks all along, the moment of buoyancy about after block =
moment of weight about after block. This equals moment of
buoyancy about after block when floating freely.
Hence we place a profile of the ship on the line of blocks, and
draw a series of waterlines parallel to the keel. For each of these
calculate the displacement and the longitudinal C.B. Draw out on
a scale of draught a curve giving the moment of buoyancy about
the after block. Where this crosses the constant line of the
moment of weight about after block will give the draught at
which the ship will ground, and so the displacement. This
deducted from the original displacement gives the pressure on the
blocks, and from the above the stability under these conditions can
be determined. In a ship with small metacentric height and large
trim by the stern, we have a combination of circumstances which
would probably cause instability. The course to pursue is to keep
the ship under control while any weight is taken by the blocks.
2. Approximately. Suppose the vessel trims / feet by the stern,
Appendix. 419
and let the after block be b feet from the centre of flotation. When
the vessel is floating* freely, imagine a force Q is applied at the
after block just sufficient to bring the vessel to an even keel.
Q = I2 ' , where M is moment to change trim I inch.
The upward force Q will decrease displacement, and the
mean draught is reduced by + T~^r ^ eet T being tons per
inch. Owing, however, to the change of trim, the mean draught is
increased by j f eet > where the centre of flotation is c feet abaft
amidships. If x is the draught at fore end when floating freely,
then the mean draught when just grounding is
. t .c M./
Theory of the Integrator. This instrument, shown in dia
gram in Fig. 79, gives by using suitable multipliers to the results
obtained
Ci) Area of a closed figure,
(li) Moment of a figure about a given axis,
(iii) Moment of inertia of figure about a given axis,
by tracing out the boundary of the figure with the pointer of the
instrument.
*cr N
Sx'
FIG. 155.
In Fig. 155 let M be the closed figure and AB the axis, P is a
pointer at the end of an arm PC which is rigidly attached to a
circle CL. The centre C of this circle is constrained to move
along the line AB. Gearing with L is another circle N, centre D,
CD always being perpendicular to AB. At the end of a radius
DE of the circle N is a recording wheel capable of rotating about
420 Appendix.
DE, and this wheel can only record movements perpendicular
toDE.
Suppose the ratio of the circles L and N is as n : i. Then
for an angular movement 6 of L the wheel radius DE will move
through nd, and if when PC is on AB, DE is at an angle a, then
when PC is at 0, DE is at an angle < = n6 + a.
In going from P to the consecutive point P' on the curve,
separated by a longitudinal interval 8.*, we have to consider the
influence of the recording wheel of two separate movements of
PC, viz.
(i) that due to the angular motion of PC ; and
(ii) that due to the horizontal transfer 5^' of the centre C
along AB.
Consider now the influence of these two movements on the
wheel
(i) Since the curve is a closed curve the net result of the
angular movement is zero.
(ii) The recording wheel moves $x' parallel to AB, and the
record on the wheel, i.e. its movement perpendicular to
DE, is 5^. cos 0,
= 5r'. cos (n6 + a),
and the total record
= /cos (nB + a) . dx 1
But 5.r = 5.r' + CP . 50 . sin 6.
Hence the total record
= /cos (nd + o) (8r  CP . sin 6 . d&]
= /cos (nB + a) dx. /CP . cos (n& + a) . sin 6 . d&.
CASE I. Take n  i, a =
The reading is /sin 6 . dx /CP . sin 2 . d& .
The second term vanishes for a complete circuit, and since
ordinate of the curve is CP sin 6, the reading is proportional to
jy . dx, and therefore to the area.
CASE 2 Take n = 2, a = o.
The reading is /cos 26 . dx + vanishing terms,
= /(i 2 sin 2 6) dx
= fdx  2 /sin 2 6 . dx
 ^ jy 2 . dx, since \dx
Appendix. 421
i.e. the reading is proportional to the moment of the area about
AB.
CASE 3. Take n 3, a  .
The reading is Jsin 3$ . dx 4 vanishing terms
= J(3 sin 6 4 si
the first term of which is proportional to area and the second
term to the moment of inertia.
Case 3 is little used in ship work. The student on first taking
up the use of the integrator is advised to take simple geometrical
figures of which the exact area and moment are known, and by
this means the accuracy of the instrument may be tested, and, if
necessary, any corrections made.
MISCELLANEOUS EXAMPLES.
1. The tons per inch immersion in salt water at a ship's waterplanes
are as follows, commencing with the L.W. P. : 12*9, 12*4, 11*5, 10*2, 8  o,
6*0, 2*2. The first five waterplanes are 21 inches apart, and the last three
are loj inches apart, the draught being 9 feet to bottom of flat keel.
(a) Determine the displacement and the vertical position of the centre
of buoyancy to the first three waterplanes.
(b) Estimate the displacement of the vessel when floating at a draught
of 10 feet I J inches in water of which i cubic foot weighs 63! Ibs.
Ans. (a) 1063 tons, 377 feet below L.W.L.
797 473
545 572
(b) 1228 tons.
2. Construct a formula giving the additional displacement, due to
I foot greater trim aft as compared with the normal trim, in terms of the
tons per inch immersion, length between draughtmarks, and the distance
of the centre of flotation abaft midships.
The vessel in question, No. I, whose normal draught is 9 feet on an
even keel, floats in salt water at a draught of 8 feet 7 inches forward and
9 feet 10 inches aft. Estimate the displacement in tons, the centre of
flotation being 7 feet abaft amidships, and the length P.P. 250 feet (draught
marks at perpendiculars).
Ans. 12 =^, i loo tons.
LJ
3. The vessel in question No. I floats at a mean draught of 9 feet
6^ inches in salt water. While in this condition she is inclined, two
plumbbobs 10 feet long being employed. The following deflections are
observed :
422 Appendix.
Forward. Aft.
3 tons through 23 ft. P to S 3 '6" ... 35"
6 7* I 5" 7'05"
Weights restored, ship came to upright.
3 tons through 23 ft. S to P 3'55" ... 3'6"
6 ,, ,, ,, 7' 1 5" " 7' 1 "
Estimate the metacentric height at the time.
Ans. 2 feet.
4. A vessel of box form, 150 feet long and 25 feet broad, floats at an
even draught of 8 feet, and has a watertight deck 8 feet above keel. If
a central compartment, 30 feet long, bounded by two transverse bulkheads
extending up to the deck, is bilged, what will be (l) the new draught of
the vessel ; (2) the alteration of the metacentric height ?
Ans. 9' 8J" nearly ; increase nearly I foot.
5. A body with vertical sides, the plan being an isosceles triangle 150
feet long and 30 feet broad at tLc stern, floats in salt water at a constant
draught of 10 feet. Determine the displacement when floating at a
draught of 9 feet 6 inches forward, 10 feet 6 inches aft
(a) by using formula obtained in question (2) above ;
(6) by direct calculation, thus verifying (a).
6. Obtain a rule for finding the area between two consecutive equi
distant ordinates of a curve when three are given. Show that the rule.
when used with levers, results in a moment error of X (intercept between
whole curve and chord), where h is the common interval.
7. The halfordinates of the waterplane of a ship 320 feet long and of
9500 tons displacement are 1*0, 165, 25*0, 29*0, 304, 306, 30*5, 29*8,
28*1, 24*1, and 15*1 feet. Find the sinkage of the vessel on passing from
the Nore to the London Docks (63 Ibs. to cubic foot).
Ans. 3*9 inches.
8. If the vessel in the last question draws F 24' 3", A 27' 9" when at
the Nore, find her draughts forward and aft when in the docks, the centre
of buoyancy being 5*1 feet abaft middle ordinate and n feet below the
centre of gravity.
Ans. F 24' 7*", A 28' oj".
9. A cigarshaped vessel with circular sections floats in salt water with
its axis in the surface. The semiordinates of the waterplane, 20 feet apart,
are, commencing from forward, o, 3, 6, 8, 7, 4, I feet respectively.
Find (i) Tons per inch immersion.
(2) Displacement in tons.
(3) Position of C.F from after end.
(4) Position of C.B. ,,
(5) Transverse BM.
(6) Position of C.B. below W.L.
Ans. (I) 276 tons; (2) 1577 tons; (3) 57 feet; (4) 567 feet;
(5) 284 feet ; (6) 284 feet.
Note. Some consideration should be given as to the simplest method
of doing this question ; (6) should be inferred from (5).
10. A vessel of constant triangular section is 245 feet long, 30 feet
broad at the waterline, and floats at 12 feet draught with vertex downwards.
Appendix. 423
When a weight of 8 tons is moved 30 feet across the deck, a shift of
8 inches is caused on the bob of a lOfeet pendulum. Find the position
of the vessel's centre of gravity.
Ans. 1 7 '64 feet from base.
n. Assuming that a barge is of uniform rectangular section, 70 feet
long and 20 feet broad, construct the metacentric diagram to scale for all
draughts between 2 feet and 10 feet; state the draught for which the
height of the metacentre above the keel is lowest, and show that in this
condition the metacentre is in the corresponding waterplane.
Ans. 8' 2".
12. A right circular cone is formed of homogeneous material, and the
tangent of the semivertical angle is O'5. Show that this cone will float in
stable equilibrium with vertex down in fresh water so long as the specific
gravity of the material is greater than 0*5 1 2.
13. A long triangular prism of homogeneous material having the same
section as the above floats in fresh water with vertex down. Show that
it will float in stable equilibrium so long as the specific gravity of the
material is greater than 0*64.
14. A lighter has a constant section 16 feet at the base, 20 feet across
the deck,iand 10 feet deep. She floats in river water 35*7 cubic feet to the
ton at a constant draught of 8 feet, the length being 80 feet. The C.G.
when laden to this draught is 6 feet above the base.
Determine the angle of heel caused by taking 5 tons of the cargo out,
this cargo being at 6 feet from the base and 6 feet from middle line.
Ans. 2j to 2\ degrees.
15. What relation exists between the transverse and longitudinal
stability of a wholly submerged body ?
Discuss the question of submarine navigation from the point of view of
longitudinal stability.
1 6. A lighter with vertical sides is 132 feet long and 30 feet broad for
a length amidships of 80 feet. The ends are formed of four circular arcs
of 30 feet radius. The draught is 10 feet, and the C.G. at this draught is
7^ feet from the bottom. Determine the metacentric height.
Ans. 435 feet.
17. Prove the rule for the distance of the centre of gravity of a hemi
sphere of radius a from the bounding plane, viz. jj . a, by finding the BM
of a sphere floating with its centre in the surface of the water. (See
question 17, p. 141.)
1 8. A ship of length 320 feet, breadth 50 feet, mean draught 19 feet,
has a displacement of 4400 tons. The tons per inch at the L.W.L. is
27, BM is 1 1 feet, and GM is 25 feet.
It is proposed to design on similar lines a ship with the dimensions
length 330 leet, breadth 51 feet, mean draught 19! feet. If G is the same
distance above the keel in both ships, what value of GM would you
expect in the new ship ?
(Use approximate formula on pp. 66 and in.) Ans. 2f to 3 feet.
19. A vessel of 700 tons displacement has a freeboard to the upper
deck of 6 feet. The C.G. is i foot above water, and the metacentre locus
is horizontal. A sea breaking over the bulwarks causes a rectangular area
424 Appendix.
50 feet long and 20 feet wide on the upper deck to be covered with water
to a depth of I foot. Calculate the loss of metacentric height
Ans. 15 foot.
20. Show that for a vessel wallsided in the neighbourhood of the
waterline, GZ = (GM + JBM tan 2 0) sin at the angle of heel 9.
Use this formula to determine the metacentric height in the upright
condition of a boxshaped vessel, 200' X 35' X 10' draught, which is found
to loll over to an angle of 5 (see p. 173). Ans. 004 foot.
21. A tank, extending across an oilcarrying vessel, is 35 feet wide, 40
feet long, and 10 feet deep. It has an expansion trunk at the middle line
4 feet wide and 6 feet long. The vessel has a displacement of 2000 tons
in salt water, and a GM of 2$ feet, the C.G. being 10 feet above the
bottom of the tank.
Find the virtual metacentric height when the tank is half full and also
when filled. The density of the oil is 0*8 as compared with sea water,
and the metacentric curve is horizontal.
Ans. (i) 154 foot ; (2) 319 feet.
22. y lt y v y tt y^ y s , and y t are six consecutive equidistant ordinates
of a plane curve : obtain the following expression for the area A of the
curve lying between y l and y t , h being the common interval :
A =
23. A foreign vessel, whose form is not known, has a certain draught
at the Nore, the seawater there being 64 Ibs. per cubic foot. Off Green
wich, the water there being 63 Ibs. per cubic foot, it is noted that when
loo tons have been unshipped the draught of water is again what it was
at the Nore. What is the seagoing displacement of the vessel ?
Ans. 6300 tons.
24. A vessel 60 feet broad at waterline has the transverse metacentre
12 feet above C.B., the latter being 10 feet below water. Find the height
of metacentre above this waterline when
(a) The beam is increased to 62 feet at the waterline, and in this
ratio throughout, the draught being unaltered ;
() The breadths at waterline are increased as above, but the lines
fined so as to maintain the original displacement and to raise the
C.B. 04 foot. Ans. (a) 2'8 feet ; (b) 364 feet.
25. In a vessel whose moment to change trim one inch is M, tons per
inch is T, and centre of flotation from after perpendicular is e times the
length between perpendiculars, show that the position for an added weight
such that the draught aft shall remain constant is  feet forward of the
centre of flotation, and thus if the C.F. is at midlength this distance is
M
2 . feet, or, approximately, oneninth the length in a ship of ordinary
form.
26. Show that the distance forward of the after perpendicular at which
a weight must be added so that the draught aft shall remain constant is
given by moment of inertia of waterplane about the A. P. divided by the
moment of waterplane about the A. P.
27. A long body of specific gravity 0*5 of homogeneous material floats
in fresh water, and has a constant section of the quadrant of a circle of
10 feet radius. Determine the metacentric height when (a) corner upwards,
Appendix. 425
(b) corner downwards, and (c) when between the positions (a) and (b).
Draw the general shape of the siability curve from zero to 360, starting
with the body corner upwards. (The C.G. of the quadrant is times the
radius from each of the bounding radii. Positions of stable and unstable
equilibrium occur alternately.)
Ans. (a} +233 feet ; (b) +233 feet ; (c) 036 foot.
28. Two ships of unequal size are made from the same model. Prove
that at the speed at which the resistance varies as the sixth power of the
speed, the same effective horsepower is required for both ships at the same
speed.
29. A vessel 375 feet between perpendiculars is designed to float at
21 feet F.P., 23 feet A. P. At this draught the displacement is 6500 tons
salt water, tons per inch 45, and centre of flotation 13$ feet abaft amidships.
The draught marks are placed on the ship 25 feet abaft the F.P. and
35 feet before the A. P. respectively. Estimate as closely as you can the
displacement when the draught marks are observed at the ship, 19' 6"
forward, 23' 10" aft, when floating in water of which 357 cubic feet
weigh I ton.
Ans. 6293 tons.
30. H.M.S. Pelorus is 300' X 36$' X 13$' mean draught, 2135 tons
displacement, and requires 7000 I.H.P. for 20 knots. On this basis
estimate the I.H.P. required for a vessel of similar form, 325' X 40' X 15!'
mean draught, 3000 tons displacement, at 21 knots speed. State clearly
the assumptions you make in your estimate.
Among others the following assumptions are made :
(1) For increased displacement caused by bodily sinkage, the I.H.P. varies as
displacement for the same speed.
(2) At the speeds mentioned in question, the I.H.P. is varying as the fourth power
of the speed.
Ans. About 10,600 I.H.P.
31. The following formula has been proposed for the E.H.P. of a
vessel at speed V knots, viz.
E.H.P. = ^ /. S . (V) 2 " + b . ( ^ V'J
where S = wetted surface in square feet.
W = displacement in tons.
L = length in feet.
/ = a coefficient for surface friction.
b = a coefficient varying with the type of ship.
A vessel 500' X 70' X 26^', draught 14,000 tons, is tried at progressive
speeds, and the curve of I.H.P. on base of speed shows the following
values, viz. at 10, 12, 14, 16, 18, 20 knots, the I.H.P. is 1800, 3100,
5000, 7500, 11,000, 15,500 respectively.
Assuming the above formula to correctly give the E.H.P., determine
the propulsive coefficients at the six speeds given.
(Take/= 0009, * = ' 2 S = 1 S'SJW x L as P 262 ')
Ans. (10) 464%; (12) 47'i / ; d4) 47'3/ ; (i6) 4 85 / ;
(i8)48'9/o5 (20)498%.
32. Using the above formula for E.H.P. (with /= 0009, b = o'2s),
determine the I.H.P. for speeds of 18 and 19! knots respectively of a
Appendix.
vessel 350' x 53^' x 20' x 5600 tons, using propulsive coefficients of
45 / and 47* / respectively.
Ans. 7660 I.H.P. ; 9740 I.H.P.
33. Draw out the metacentric diagram for all draughts of a square log
of 2 feet side, floating with one corner down.
Supposing the log to be homogeneous, determine the limits between
which the density must be in order that it shall float thus in stable equi
librium in fresh water.
Ans. Between 0*28 and 0*72.
34. A rectangular vessel is 175 feet long, 30 feet broad, 20 feet deep,
and floats at a draught of 8 feet, with a metacentric height of 5 feet. Find
the draught forward and aft, and the metacentric height due to flooding
an empty compartment between bulkheads 120 feet and 150 feet from the
after end. Ans. F. 13' 4!" ; A. 6' 7f" ; 42'.
35. In a wallsided vessel, show that for an angle of heel 6 the co
ordinates of the C.B. referred to axes through the C.B. in the upright
condition are x = BM, .tan 9 ; y = $BM . tan 2 6. (BM, refers to the
upright condition.)
36. Using the above, show that a wallsided vessel will heel to angle 6
by shifting a weight w a distance d across the deck, 6 being given by the
equation
the suffix o referring to the upright condition.
Thus, for a zero metacentric height the heel 9 is given by
Y wxd
tan = 2 .
37. Show that a wallsided ship having an initial negative metacentric
height will heel to an angle of 9 = tan. / 2 GM Q, and will then have a
metacentric height of 2GM . /i + 2 GM = 2 ^^.
/ \/ BM ' cos 6
38. Prove (by using BM =  J that the C.G. of a segment of a circle
radius a, subtending an angle of 26 at the centre, is distant from the
centre a. :  : r  ^, and thus for a semicircle ( =  ) the C.G.
3 v sin cos y 2j
is  from centre. (Area of segment is area of sector less area of
triangle, or a*0 a 2 sin 6 cos 6.)
39. A vessel of 300 feet length floats at a draught of 12 feet forward,
15 feet aft. (Tons per inch 18 ; moment to change trim I inch, 295 tons
feet ; C.F. 12 feet abaft midships.) It is desired to bring her to a draught
not exceeding 12 feet forward and aft. How could this be done?
Ans. Remove 350 tons 42^ feet abaft amidships (account is taken
of increase of mean draught due to change of trim. )
40. Draw a curve of displacement for all draughts of a cylindrical
vessel of diameter 20 feet and 150 feet long, and find, by using the curve,
the distance of the C.B. from the base when floating at a draught of
15 feet.
Appendix. 427
41. Draw the curve of displacement of a vessel of 14 feet draught
having the following displacements up to waterlines 2 feet apart, viz.
2118, 1682, 1270, 890, 553, 272, 71 tons, and by it find the position of
the C.B. with reference to the top waterline. Suppose the tons per inch
is 18*56, check your result by Morrish's formula.
Ans. 545 feet.
42. Prove the rule given on p. 19 for the volume of a sphere, by using
Simpson's rules at ordinates, say,  the radius apart.
(An exact result should be obtained, because the curve of areas is a
parabola, which Simpson's rule correctly integrates.)
43. A boxshaped vessel 140 feet long, 20 feet broad, 10 feet draught
is inclined by shifting 7 tons 15 feet across the deck, and heels to an angle
tan 1 (J). Find the metacentric height (a) accurately, (b) by ordinary
method.
Ans. (a) 042 foot, (b) 0*52 foot.
44. A long iron pontoon, of section 6 feet square and of uniform
thickness, floats when empty in seawater, but lolls over in fresh water.
Find the thickness of the iron. (When the M curve is at middepth the
draughts are 1268 foot and 4732 feet. Of these the curve drops for
increase of draught only in the former, so that for increase of draught as
occurs in fresh water there is a negative metacentric height).
Ans. J inch.
45. A solid is formed of a right circular cylinder and a right circular
cone of the same altitude h on opposite sides of a circular base radius r.
It floats with the axis vertical, the whole of the cone and half the cylinder
5 y4 21
being immersed. Prove that the metacentric height is '~T~jr**
so that for stable equilibrium r must be greater than 0*934^.
46. A shallowdraught lightly built vessel is being launched. State
the nature of the strains on the structure that will be experienced as she
goes down. (From this point of view, the practice of some firms in
launching torpedoboat destroyers is of interest.)
47. A vessel has a list to starboard due to negative metacentric height
when upright. It is found that the addition of weights in the 'tween decks
on the port side increases the list to starboard. How do you explain this ?
48. If a swan or duck is floating in a pond, and reaches down to the
bottom for food, why does the bird find it necessary to work with her feet
to keep the head down and the tail up ?
49. When floating in water, why is it necessary to keep the arms
below? If the arms are raised out, what happens, and why?
50. If a certainsized tin is placed in water it will not float upright.
When a certain quantity of water is poured in it floats upright in stable
equilibrium. State fully the conditions of stability which lead to this
result, bearing in mind the large loss of metacentric height due to the free
surface of the water inside.
51. A cube 12 inches side weighs 104 Ibs. Investigate the stability
in fresh water, (a) with two faces horizontal, (b) with two faces only
vertical and one edge downwards, (c) with a corner down.
52. In going through the Caledonian Canal, the writer has noticed
that the level of the water falls amidships. How do you account for this?
53. If the water in question 36, Chap. III., goes right away with the
tide, and the mud is very deep, investigate the stability of the vessel,
the original metacentric height being 4 feet.
Ans. Negative GM of foot.
428
Appendix.
54. In a boxshaped vessel, 200 feet long, 30 feet wide, 10 feet draught,
and having its C.G. li'Sfeet above the keel, a central transverse com
partment 50 feet long (assumed empty) is opened up to the sea. Will
this vessel be stable after damage and free from danger in still water with
a row of sidelights open, the lower edges of which are 1 5 feet above the
keel?
The GM when intact is 07 foot, and when damaged is O'5 foot, and
new draught is 13,} feet, and in the final condition the vessel is all right, but
there are intermediate conditions to consider, and the following table gives
results of calculations with various depths of water in the middle com
partment :
Height of water in
compartment in !eet.
I
2
3
6
Draught of
water in feet.
Metacentric height
in feet
093
071
032
croi
+o'34
+05
Thus in the early stages the metacentric height is negative, and the
ship will "loll," and if the hole through which the water enters is of
comparatively small dimensions there would be an appreciable time for
the list to develop.
The above is taken from Prof. Welch's paper from the N.E. Coast
Institution, 1915, on "The Time Element and Related Matters in some
Ship Calculations," to which the reader is referred for a further develop
ment of the subject.
SOLUTION OF QUESTION No. 21, CHAP. II., AND No. 36, CHAP. III.
The author has had a number of requests as to the solution of these
examples, and as they illustrate an important principle the solution is given
below.
FIG. 156.
The metacentre when floating freely is readily obtained, viz. 16 feet
from the base line.
When the water level sinks 6 feet, the lower portion sinks into the
mud, say x feet. Then, since the mud has a s.g. of 2, we take in the area
Appendix. 429
Owl twice. We equate the new displacement to the old, or
(x + 6) 2 + * 2 = 144
from which x = 4*94 feet.
For a small inclination one half the buoyancy of the portion Ow/will
act through m the metacentre of Owl, and the buoyancy of OWL' will
act through M' the metacentre of OW'L', the portion Owl then being
included twice. Om = 6'59' and OM' = 14*59'. The total buoyancy will
act through a metacentre M such that
(io'94 2 x I4'59) + (4'94 2 X 659) = I2 2 x OM,
from which OM = 13 '2 feet.
That is, the new metacentre is 2 '8 feet below the original metacentre, and
as the C.G. of the ship has not been affected, the loss of metacentric height
is 2 feet, about.
APPENDIX B
TABLES OF LOGARITHMS, SINES, COSINES,
AND TANGENTS, SQUARES AND CUBES.
LOGARITHMS. For some calculations considerable trouble is
saved by using logarithms. One instance of this has been already
given on p. 317. A table of logarithms is given on pp. 434, 435,
to four places of decimals, which gives sufficient accuracy for
ordinary purposes. To the right of the table are given the
differences for I, 2, 3, etc., which enables the logarithms of
numbers of four figures to be obtained.
Thus log 2470 = 3*3927. The decimal part is obtained from
the table, the whole number being 3, because 2470 is between 1000
and 10,000 (log 1000 = 3, log 10,000 = 4). The log of 2473 is
obtained by adding to the above log the difference in the table for
3, viz. 5, i.e.
log 2473 = 3*3932.
The following are the principal relations in logarithms, viz. :
log (M x N) = log M + log N
log (jj) = log M logN
log (M) w = n . log M
log VM =  . log M
Thus multiplication is turned into addition, division is turned
into subtraction, the raising to a power is turned into multiplication,
and the taking of a root is turned into division.
The decimal portion of a logarithm is always kept positive, and
43 2 Appendix.
the following are the values of the logarithm of the number 239
for various positions of the decimal point :
log 23,900 = 43784
log 2,399 = 33784
log 239 = 23784
log 239 = 13784
log 239 = 03784
log 0239 =  i + 03784 = [3784
log 00239 =  2 + 03784 = 23784
log 000239 =  3 + o'3784 = 3'3784
Example. To find the cube root of 10*75 :
log 1075 = 1*0315
log t/(iQ75) = J (10315)
= 03438
03438 = 03424 + 00014
.*. V 10 '75 = 2*207
Example. To find the value of (5725)$:
log 5725 = 07578
log (5725)* = f (07578)
= 26523
/. (5725)2 = 449'i
Example.!* find the value of <9 2 30 * > ('4'o8)
5267
log (9*31)1 = (39652)
= 26435
log (i4o8) 3 = 3 (11485)
= 34455
log 5267 = 37216
.. Iog [(9*3.)3x( 14 o8)*j = 2 6435 + 34455  376
The number of which this is the log is 233
. (9231)* x (I4'o8) 3 =
5267
Example. Find the value of 512 x 50*5 x 0*0037.
log 512 = 27093
log 505 = 7033
log 00037 = 35682
log (product) = 19808
or product required = 95 66
Appendix. 433
Example. Find the value of Vo'oo765.
log 0*00765 = 3*8837
= 4 + 18837
log Vo00765 = i + 0*4709
/. Vo'00765 = 0*2957
NAPIERIAN OR HYPERBOLIC LOGARITHMS.
These logarithms, which are also termed "natural," are
calculated to the base e  2718, and the following relation holds :
log.N = 2*3 log 10 N
Ordinary logarithms are calculated to the base 10.
TABLE OF SINES, COSINES, AND TANGENTS.
On pp. 436, 437, is given a table showing the values of the
trigonometrical ratios, sines, cosines, and tangents, of angles up
to 90, to three places of decimals, which will be found sufficiently
accurate for ordinary purposes.
2 F
LOGARITHMS.
1
2
3
4
5
6
7
8
9
1 2 34 5 6\7 8 9
10
11
12
13
14
0000
0414
0792
1139
1461
0043
0453
0828
"73
1492
0086
0492
0864
1206
1523
0128
0531
0899
1239
1553
0170
0569
0934
1271
1584
021202530294
0607106450682
0969100411038
I303I335I367
1614 1644 1673
03340374
07190755
1072 1106
1399 1430
17031732
4 8 12:17 21 25
4 8 ii!i5 19 23
3 7 1014 17 21
3 6 10 13 16 19
3 6 912 15 18
29 33 37
26 30 31
24 28 31
23 26 29
21 24 27
15
16
17
18
10
1761
2041
2304
2553
2788
1790
2068
2330
2577
2810
1818
2095
2355
2601
2833
1847
2122
2380
2625
2856
1875
2148
2405
2648
2878
1903 1931
2175 22OI
24302455
2672 2695
2900 2923
1959
2227
2480
2718
2945
19872014
2253 2279
25042529
2742 2765
2967 2989
368
3 5 8
257
2 5 7
247
n 14 17
ii 13 16
10 12 15
9 12 14
9 " 13
20 22 25
18 21 24
17 20 22
16 19 21
16 18 20
20
21
22
23
24
25
26
27
28
20
30
31
32
33
34
3010
3222
3424
3617
3802
3032
3243
3444
3636
3820
3054
3263
3464
3655
3838
3075
328 4
3483
3674
3856
3096
3304
3502
$692
3874
3Il8
3324
3522
37"
3892
3139
3345
354i
3729
3909
3160
3365
356o
3747
3927
3181 3201
3385 3404
35793598
37663784
3945 3962
246
2 4 6
2 4 6
2 4 6
245
8 ii 13
8 10 12
8 10 12
7 9 "
7 9 ii
15 17 19
14 16 18
M 15 17
13 15 17
12 14 16
3979
4150
43M
4472
4624
3997
4166
4330
4487
4639
4014
4i8^
4346
4502
4654
4031
4200
4362
4518
4669
4048
4216
4378
4533
4683
4065
4232
4393
4548
4698
4082
4249
4409
4564
4713
4099
4265
4425
4579
4728
4"64I33
4281 429
4440 4456
45944609
47424757
2 3 5
235
2 3 5
2 3 5
i 3 4
7 9 10
7 8 10
689
689
679
12 I 4 15
II 13 15
II 13 14
II 12 14
10 12 13
4771
4914
5051
5185
5315
4786
4928
5065
5198
5328
4800
4942
579
5211
5340
4814
4955
5092
5224
5353
4829
4969
5105
5237
5366
4843
4983
5"9
5250
5378
4857
4997
5132
5263
5391
4871
5011
5H5
5276
5403
48864900
50245038
5I595I72
5289 5302
54165428
I 3 4
i 3 4
3 4
3 4
3 4
679
678
5 7 8
568
568
10 ii 13
10 II 12
9 II 12
9 10 12
9 10 ii
35
36
37
38
30
5441
5563
5682
5798
59"
5453
5575
5 6 94
5809
5922
5465
5587
5705
5821
5933
5478
5599
5717
5832
5944
5490
5611
5729
5843
5955
5502
5623
5740
5855
5966
5514
5635
5752
5866
5977
5527
5647
5763
5877
5988
5539
5658
5775
5888
5999
5551
5670
5786
5899
6010
I 2 4
I 2 4
I 2 3
I 2 3
I 2 3
5 6 7
5 6 7
5 6 7
5 6 7
457
9 10 ii
8 10 ii
8 9 10
8 9 10
8 9 10
40
41
42
43
44
6021
6128
6232
6335
6435
6031 6042
61386149
6243 6253
6345 6 355
6444 6454
6053
6160
6263
6365
6464
6064
6170
6274
6375
6474
60756085
6i8o!6i9i
62846294
6385;6395
64846493
6096
6201
6304
6405
6503
6107
6212
63H
6415
6513
6117
6222
6325
6425
6522
I 2 3
I 2 3
i 2 3
i 2 3
I 2 3
4 5 6
456
4 5 6
456
4 5 6
8 9 10
7 8 9
7 8 9
789
7 8 9
45
46
47
48
40
6532
6628
6721
6812
6902
65426551
66376646
6730 6 739
68216830
69116920
6561
6656
6749
6839
6928
657i
6665
6758
6848
6937
65806590
6675 6684
6767 6776
6857 6866
69466955
6599
6693
6785
f75
6964
6609
6702
6794
6884
6972
6618
6712
6803
6893
i 2 31 4 5 6
i 2 3  4 5 6
i 2 3  4 5 5
123445
i 23445
7 8 9
7 7 !
678
678
678
50
51
52
53
54
6990
7076
7160
7243
7324
6998
7084
7I68 1
7251
7332
7007
7093
7U7
7259
7340
7016
7101
7185
7267
7348
7024
7110
7i93
7275
7356
7033 7042
7118 7126
72027210
7284 7292
7364 7372
7050
7135
7218
7300
7380
7059
7H3
7226
7308
7388
7067
7152
7235
73i6
7396
i 23345
123345
122345
i 22345
122345
678
678
6 7 7
667
667
LOGARITHMS.
1
2
3
4
5
6
7
8
9
123
456
789
55
56
57
58
59
7404
7482
7559
7634
7709
7412
7490
7566
7642
7716
7419
7497
7574
7649
7723
7427
7505
7582
7657
773i
7435
7513
7589
7664
7738
7443
7520
7597
7672
7745
7451
7528
7604
7679
7752
7459
7536
7612
7686
7760
7466
7543
7619
7694
7767
7474
755
7627
770
7774
2 2
2 2
2 2
3 4 5
3 4 5
3 4 5
344
344
5 6 7
5 6 7
5 6 7
5 6 7
5 6 7
60
61
62
63
64
7782
7853
7924
7993
8062
7789
7860
793 i
8000
8069
7796
7868
7938
8007
8075
7803
7875
7945
8014
8082
7810
7882
7952
8021
8089
7818
7889
7959
8028
8096
7825
7896
7966
8035
8102
7832
7903
7973
8041
8109
7839
7910
798o
8048
8116
7846
7917
7987
8055
8122
2
344
344
334
334
334
566
5 6 6
5 6 6
5 5 6
5 5 6
65
66
67
68
69
8129
f'95
8261
3325
8388
8136
8202
8267
8331
8395
8142
8209
8274
8338
8401
8149
8215
8280
8344
8407
8156
8222
8287
8351
8414
8162
8228
8293
8357
8420
8169
8235
8299
8363
8426
8176
8241
8306
8370
8432
8182
8248
8312
8376
8439
8189
8254
8319
8382
8445
334
334
334
334
234
5 5 6
5 5 6
5 5 6
456
456
70
71
72
73
74
8451
8513
8633
8692
8457
8519
*579
8463
8525
8585
8645
8704
8470
8531
8591
8651
8710
8476
8537
8597
8657
8716
8482
8543
8603
8663
8722
8488
8549
8609
8669
8727
8494
fl 55
8615
8675
8733
8500
8561
8621
8681
8739
8506
8567
8627
8686
8745
234
234
234
234
234
456
455
4 5 5
455
4 5 5
75
76
77
78
79
8751
8808
8865
8921
8976
8756
8814
8871
8927
8982
8762
8820
8876
8932
8987
8768
8825
8882
8938
8993
8774
8831
8887
8943
8998
8779
8837
8893
8949
9004
8785
8842
8899
8954
9009
8791
8848
8904
8960
9015
8797
8854
5910
8965
9020
8802
8859
8915
8971
9025
I 2
2
2
2
2
233
233
233
233
233
4 5 5
4 5 5
445
445
445
30
31
32
J4
9031
9085
9138
9191
9243
9036
9090
9H3
9196
9248
9042
9096
9149
9201
9253
9047
9101
9154
9206
9258
9053
9106
9159
9212
9263
9058
9112
9165
9217
9269
9063
9117
9170
9222
9274
9069
9122
9175
9227
9279
9074
9128
9180
9232
9284
9079
9133
9186
9238
9289
2
2
2
2
2
233
233
233
233
233
445
445
445
445
445
J5
J6
(8
19
9294
9345
9395
9445
9494
9299
9350
9400
945
9499
9304
9355
9405
9455
9504
9309
9360
9410
9460
959
9315
9365
9415
9465
9513
9320
9370
9420
9469
9518
9325
9375
9425
9474
9523
9330
938o
9430
9479
9528
9335
9385
9435
9484
9533
9340
9390
9440
9489
9538
2
2
233
235
223
223
223
4 4 5
4 4 5
3 4'4
344
344
'0
'2
3
4
9542
9590
9638
9685
9731
9777
9823
9868
9912
9956
9547
9595
9643
9689
9736
9782
9827
9872
9917
9961
9552
9600
9647
9694
974i
9786
9832
9877
9921
9965
9557
9605
9652
9699
9745
9791
9836
9881
9926
9969
9562
9609
9657
J703
9750
9795
9841
9886
9930
9974
9566
9614
9661
97o8
9754
9800
9845
9890
9934
9978
957i
9619
9666
9713
9759
9805
9850
9894
9939
9983
9576
9624
9671
9717
9763
9809
9854
9899
9943
9987
9628
9675
9722
9768
9814
9859
9903
9948
9991
9586
9633
9680
9727
9773
223
223
223
223
223
3 4 A
344
344
344
344
344
344
344
344
334
5
6
7
8
9
9818
9863
9908
9952
9996
223
223
223
223
223
8N to r> ON M Tf vo oo  1 co vo r^ ON i cotor^oo O '
ONOO r^vo VO vo^frcorort  O ON ONOO t^VO vo vo
Tf vo r^. ON N ,3?*^ *"* ^
vo O to ON co t> >* to
<O t^. l>. t^OO OO ON ON
OOOOOOOOOOOOOOOOO
b "
I^O Tft^O'COVOOQ w Tfvo ON'
N PO CO CO ^f i rf Tf Tf 10 vo vp vov
N t^. Tf N M 11 COVO O ^
COVO O Tf OO N VO O vo ON Tf O
t^ t^oo oo oo ON ON p p p r
MMMM^MMCtMCICI
Tf Tf ONOO N 11 VO t^VO N
M O OO 1^ t^ t^ t^OO O CO
>0 ON ON O ** N co Tf vo t^
00 ONpNp
N W N cococococococo
oft ft ft
ft ft ft ft ft
ft ft ft ft ft
\O vo vO vo vo vo vO vo vO
t^ ^00 OO ON ON O
ON
vo
vo votOTfTfrocOM N
OJOOOOOOOOOOOOOOOOJOOO
1VO CO ONVO CO
VO vp vp vp vp
t^OO O  N CO ^vO t^OO
Tf ON to O VO f> ONVO CO O
LOVO OO O H CO TfVO OO O
OOOCX3 pO\O\O%O*O\
^ O CO CO fO co co
o cocococococococococoi
ft ft ft ft ft
ft ft ft ft ft
wvo vo t>. t^OO 00 ON ON O
cocofocomcococoro^f
QMMNNCOCOTfTftO
TfTfTfTfTjTjTfTfTfTf
ON ON ON ON ONOO OO t^vo
ON ON ON ON ON ON ON ON ON
t ON ON ON ON ON ON ON ON ON
to to Tf co ti O ONOO VO to CO N O 00
VO
vo
NON
*
CO N ONOO vo to
to CO N O O\
> ONOO vO to
O N CO ro * "^O t^.OO ONO.O N ro Tf to to
N 11 O ON t
4 N CO 9 m\O 1^ t^OO
OOOOOOOOOO
b "
uiTfcOCON M O O ONOO
ft ft ft ft ft
ft ft ft ft ft
OVO v t^ t^00 00 ON ON O
O N CO VOVO OO ON M N *
vo ! to N HI o ON ONOO r*.
NNNNNNHIHIHIHI
VO tn fO CO
O HI N Tf vovO t% ON Q
f.VO vo 3 CO N 11 Q C
poop o p p 5 p
O N ^vo 00 O N to vo
oo oo oo
vo OO ON O ! to rf
OO CO OO ON ON ON ON
ON ON ON ON ON ON ON
r^^Aj VAJ ~M w^ *^ w ^ >
gSpJsasiSio;?
t^ HI XO HI HI VO VO VOVO HI
vOicvotOHiO' l <tfaM>
OO O n to LO t^ ON HI tovo
1^ rf m O
t^nloO CO
ON tOO >* m HI t 10 ro rj
O vr>OO O fOOO ONOO S>!
totoo o^vo c^ 10^
ON O
a a a
vovo VO t^ t^OO 00 ON ON
QHIHlMNfOtOt'lW
OOQOOOOOOOQOOOOOOOOO
io*O vb t* t^QO oo ON ON O
000000000000000000 ON
,s
O ON CO>O ON N irjOO HI T^
CO N N HI O O ONOO OO t^
QvOvOVO'O^O Lr>u)iotr)
N wi t>* O N vooo O
)iO^COtON H. O O
u^vomvoxovovom
HI HI N tO fO ^~
r^ t^ t^ t^ r> t^ .
N t^ flOO CO ON rf ON rj ON
I s * t^QO OO ON ON O O M HI
ON fOOO to fx N VO
. co^^vo vovo vo
t^OO 000000 OOOOOOOOOOOOOOOOOOOO
Tf ON ^
N M to
& pp pp
ONHi coiOl^a
fOt^.HivO NOO
HI CO tJiOO O N to t^ O N
N N O N tOfOfOfOr)^
VO CO HI O O O N rfOO N
\OOO <* t^ O COVO ON CO
VO VOVO VO VO ^O t
4 1 , o a & a
4j >* *
to to to CO CO
a
vovo vo t^ t^.00 00 ON ON O
vovovotovovovovo lOVO
HI ONVO rj HI 00 VO CO
^" O t^ ^ HI t% ^ o vo
ONpNONONONpNpNpN&>
to ON vo HI _
O ON ON ONOO
ONOO 00 00 00
HI o> r^vo rf N
<X) ON O O " N CO TJ
04 M COCOCOCOCOCO
HiOOvO ^N O\tvoN O
co co TJ vovo vo t^oo ON O
^^^^<4'^Th'^'<iv
tt^VO <O "^ tO 10 T* Tj Tf
t^OO ON O n M CO Tt tovo
N N N COfOCOCOfOCOCO
r>.oo ON o >i
t^oO ON HI N ro <* vr.vo
rjrt Tj
r. co to to to to
a a &
u><O VO t^ r^OO 00 ON ON
N N N N C?C?off?e?
vovO VO t^ t~xOO OO ON ON O
McicfCtctMci'ttitcn
TABLE OF SQUARES AND CUBES OF NUMBERS
UP TO 50, RISING BY 005.
The following table has been prepared, as squares and cubes
of numbers are frequently required in ship calculations. Ordinates
usually will not be measured more accurately than to the nearest
0*05 ; in most cases the nearest decimal point is sufficiently accurate.
The squares and cubes are taken to the nearest whole number,
which is all that is necessary in ship calculations.
,
.
4
u

I
u

s
1
Cubes.
g
a
3
Cubes.
ifl
S
3
Cubes.
3
fe
0*
fc
JT
3
fe
&
005
__
__
'55
2
4
305
9
28
O'lO
60
3
4
3*10
10
30
015
65
3
4
3'!5
10
31
0'20
70
3
5
3'2o
IO
33
025
030
^_
32
3
3
1
325
330
II
II
o;35
85
3
6
3'35
II
38
90
4
7
3'4
12
39
o'45
95
4
7
345
12
41
050
2'00
4
8
12
43
0'55
205
4
9
3'55
13
45
o'6o
2'10
4
9
3'bo
13
47
065
215
5
10
365
13
49
070
2'20
5
ii
370
14
51
075
080
*~ '
225
230
5
5
ii
12
38^
H
53
55
085
2'35
6
13
385
15
57
090
240
6
14
390
15
59
o'95
I'OO
245
250
6
6
II
3'95
400
16
16
62
64
105
2'55
7
17
4 '05
16
66
10
260
7
18
410
17
69
15
2
265
7
19
415
17
20
2
270
7
20
420
18
74
25
2
2
8
21
4]25
18
30
2
2
2'8o
8
22
18
80
'35
2
2
285
8
23
4'35
19
82
40
45
2
2
3
3
2*90
295
8
9
3
440
4'45
19
20
8 5
88
2
3
300
9
27
450
20
91
Appendix.
439
I
i
Cubes.
1
I
1
1
1
Cubes.
1
fe
t
D 1
U3
Cubes.
4'55
460
465
470
$1
21
21
22
22
23
94
97
101
104
107
6'8o
685
690
6'95
700
46
47
48
48
49
314
321
329
336
343
9'05
9*10
9'i5
920
9^5
82
83
84
11
741
754
766
779
791
485
490
4'95
500
*J
24
2 4
25
25
114
118
121
125
7'05
710
715
720
7'2C
50
50
5 1
52
350
358
366
373
7 o T
o u
9'35
940
9'45
9'5o
87
88
89
90
804
817
831
844
857
505
510
5'i5
5 '20
525
26
26
27
3
28
I2 9
133
137
I 4 I
H5
/ O
7'30
7'35
740
7'45
7'5o
53
53
54
1
56
3 51
389
397
405
4 r 3
422
955
9*60
965
970
975
o'8o
9i
92
93
94
95
871
885
899
913
927
3
5'35
540
5 '45
5^
29
29
30
30
149
'53
'I 7
162
166
7'55
7 60
765
770
7*7C
11
59
59
430
439
448
457
985
990
995
10 '00
y
97
98
99
100
941
956
970
985
1,000
5'55
560
5'65
57o
575
5.o n
31
31
32
32
33
171
176
180
185
190
/J
7'80
7'85
790
7'95
800
61
62
62
63
64
5
475
484
493
502
512
1005
lO'IO
IOT5
I0'20
1025
JQ'OQ
101
1 02
103
104
lol
1,015
1,030
1,046
1,061
1,077
50
5^5
590
5'95
6'oo
34
34
35
35
36
95
200
205
211
216
805
810
815
820
82C
65
66
66
67
AQ
522
531
54i
551
Cfi2
10*35
10*40
1045
10*50
107
108
109
no
I >93
1,109
1,125
1,141
1,158
6*05
6'io
6''5
6*2O
625
6"2O
37
37
38
38
39
221
227
233
23*
244
^
830
8'35
840
8'45
850
69
70
7i
7i
72
5O2
572
582
593
603
614
!0'55
io'6o
10*65
1070
1075
10*80
III
112
"3
114
116
i,i74
1,191
1,208
1,225
1,242
635
6 40
645
650
4U
40
41
42
4 2
250
256
262
268
275
8'55
860
865
870
8'7C
73
74
75
76
625
636
647
659
10*85
1090
1095
H'OO
118
119
1 20
121
1,277
i,295
1,313
i,33i
6'55
660
665
670
675
43
44
44
45
46
281
287
294
301
308
/D
8'80
885
890
8'95
9 oo
77
77
78
79
80
81
681
693
705
717
729
1135
II'IO
11*15
1 1 20
1125
122
123
124
125
I2 7
1,349
1,368
1,386
1,405
1,424
440
Appendix.
Numbers.
t
!
Cubes.
Numbers.
1
Cubes.
Numbers
1
1
Cubes.
1130
1 1 '35
1 1 '40
1145
1150
128
129
130
131
132
,443
,462
,482
,501
,521
1355
13*60
I365
1370
1375
184
185
1 86
1 88
189
2,488
2,515
2,543
2,57i
2,600
2fi?8
I5*8o
I585
1590
I5'95
16*00
250
251
253
254
256
3,944
3,982
4,020
4,058
4,096
"'55
ir6o
11*65
1170
i33
i35
136
137
T 78
,541
,56i
,581
,602
62?
13 oO
1385
13*90
1395
14*00
190
192
193
195
196
2,657
2,686
2 ,7I5
2,744
16*05
16*10
16*15
16*20
258
259
261
262
*>f\A
4,135
4,173
4,212
4,252
75
1180
1185
1190
"'95
1200
13
139
140
142
143
144
,O32
,643
,664
,685
,706
,728
14*05
14*10
1415
14*20
14*25
197
199
200
202
203
2,774
2,803
2,833
2,863
2,894
16*30
1635
16*40
16*45
1650
266
26 7
26 9
271
272
,291
4,331
4,371
4,411
4,451
4,492
I205
1210
1215
1220
I2'2C
145
146
148
149
,750
,772
,794
,816
C 7
*4 o u
1435
14*40
1445
H50
206
207
209
2IO
xy*4
2,955
2,986
3,oi7
3,049
1655
16*60
16*65
16*70
274
276
277
279
C T
4,533
4,574
4,616
4,657
A 600
12 25
I230
I2'35
12*40
I2'45
12*50
I 5
IS'
153
154
155
156
>j
,861
,884
,907
,930
,953
H'55
14*60
!4"65
14*70
!4"75
iA*8rj
212
213
215
216
218
3,o8o
3,112
3,144
3,177
3,209
1U /i
16*80
1685
16*90
1695
17*00
282
284
286
287
289
4,742
4,784
4,827
4,870
4,913
I2'55
12*60
I2*65
12*70
158
'59
160
161
167
i,977
2,000
2,024
2,048
1485
14*90
1495
15*00
51
221
222
224
225
>^4 Z
3,275
3,308
3,341
3,375
1705
I7'io
I7I5
1720
I 7*2C
291
292
294
296
298
4,956
5,000
5,044
5,o88
51 33
12 75
12*80
12*85
12*90
12*95
13*00
^ 3
164
165
166
1 68
169
2 ,73
2,097
2,122
2,147
2,172
2,197
I505
15*10
I5I5
15*20
I525
227
228
230
231
233
3,409
3,443
3,477
3,512
3,547
? rXo
*/ *3
1730
I735
17*40
I7'45
I7*50
299
301
303
3j
306
, L O3
5,178
5,223
5,268
5,3H
5,359
13*05
1310
13*15
1320
170
172
173
174
11 f\
2,222
. 2,248
2,274
2,300
1 S 3
I535
1540
i5'45
i5'5o
234
236
237
239
240
35 52
3,6i7
3,652
3,688
3,724
1755
I7*6o
1765
IT70
I7"7 c
308
310
312
313
71 C
5,405
5,452
5,498
5,545
5CQ2
*3 2 5
I3"3o
I335
1340
1345
I350
170
177
178
1 80
181
182
2,326
2,353
2,379
2,406
2,433
2,460
1555
15*60
1565
15*70
1575
242
243
245
246
248
3,76o
3,796
3,833
3,870
3,907
/ IJ
I7*80
I785
I7*90
I795
18*00
6 1 J
3 1 7
3i9
320
322
324
oy 6
5,640
5,687
5,735
5,784
5,832
Appendix.
441
1
3
fe
I
Cubes.
1
D
!
i
Cubes.
1
&
1
Cubes.
I805
1810
1815
1820
1825
329
331
333
5,881
5.930
5,979
6,029
6,078
6 128
2030
2035
2040
2045
2650
412
414
4 i6
4 i8
420
8,365
8,427
8,490
8,552
8,615
2255
22"6O
2265
2270
2275
2?'8f>
509
5"
5i3
5i5
5i8
c?o
11,467
",543
11,620
11,697
",775
U8c?
lo 30
i835
1840
1845
1850
335
337
339
340
342
6,179
6,230
6,280
6,332
2055
20 '60
20*65
2070
422
424
426
428
8,678
8,742
8,806
8,870
8Q1A
2285
2290
2295
2300
^zu
522
524
527
529
n,930
12,009
12,088
12,167
i8'55
1860
1865
1870
i875
rC'SX
344
346
348
350
352
6,383
6,435
6,487
6,539
6,592
ft 6/1 c
20 75
2080
2085
2090
2095
2I'OO
43 1
433
435
437
439
441
,Vj4
8,999
9,064
9,129
9,195
9,261
2305
2310
2315
2320
2325
531
534
536
538
54i
12,247
12,326
12,407
12,487
12,568
15 oO
I885
iS^O
1895
I900
353
355
357
359
361
, U 45
6,698
6,751
6,805
6,859
2105
2I"IO
2115
2120
443
445
447
449
9,327
9,394
9,461
9,528
23 30
2335
2340
2345
2350
543
545
548
550
552
12,649
12,731
12,813
12,895
12,978
I905
I9IO
I9I5
I920
I925
363
3 ^ 5
367
369
37i
6,913
6,968
7,023
7,078
7,133
7I&Q
21 25
2I30
2135
2I40
2145
2150
452
454
456
458
460
462
,59
9,664
9732
9,800
9,86 9
9,938
2355
2360
2365
2370
2375
o7'Xn
555
557
559
562
564
f.f.
13,061
13144
13,228
13,312
13,396
I930
I935
I940
1945
I950
37 2
374
376
378
380
,ioy
7,245
7,301
7,358
7,415
2i'55
2160
21*65
2170
21 '7
464
467
469
47i
A.T\
10,008
10,078
10,148
10,218
2385
2390
2395
2400
569
57i
574
576
13,40!
13,566
13,652
13,738
13,824
I955
I960
I965
I970
1975
382
384
386
388
390
7,472
7,530
7,5^7
7,645
7,704
21 75
2 1 'SO
2I85
2I90
2195
22'OO
H/O
475
477
480
482
484
lUj^oy
10,360
10,432
10,503
10,576
10,648
2405
2410
2415
2420
2425
578
58i
583
586
588
I3,9H
13,998
14,085
14,172
14,261
19 oO
1985
1990
1995
2000
392
394
396
398
400
7j
7,821
7,88 1
7,940
8,000
22O5
22'IQ
2215
2220
22'?C
486
488
491
493
/IOC
10,721
10,794
10,867
10,941
24 3
2435
2440
2445
2450
59
593
595
&
! 4>349
14,438
H,527
14,616
14,706
2O'O5
2010
20 15
2020
2O25
402
404
406
408
410
8,060
8,121
8,181
8,242
8,304
22 25
2230
2235
22*40
2245
22*50
47J
497
500
502
504
.506
11,015
11,090
11,164
11,239
",3i5
n,39i
2455
2460
2465
2470
2475

608
610
613
14,796
14,887
H,978
15,069
15,161
442
Appendix.
Numbers.
rt
D
cr
M
Cubes.
Numbers.
t
m
1
Cubes.
Numbers.
t
1
Cubes.
2480
2485
2490
2495
2500
615
618
620
623
625
15,253
15,345
15,438
15,531
15,625
2705
2710
2715
2720
2725
273
2735
2740
2745
2750
732
734
737
740
743
745
748
75i
754
756
19,793
19,903
20,013
2O,I24
20,235
20,346
20,458
20,571
20,684
20,797
2930
2935
2940
29H5
2950
858
861
864
867
870
25,154
25,28 3
25,412
25'542
25,672
2505
2510
25I5
2520
2525
2530
2535
2540
2545
2550
628
630
633
635
638
640
6 43
645
648
650
15,719
15,813
15,908
16,003
16,098
16,194
16,290
16,387
16,484
16,581
2955
2960
2965
2970
2975
2980
2985
2990
2995
3000
873
876
879
882
885
888
891
894
897
900
25,803
25,934
26,066
26,198
2 ^33I
26,464
26,597
26,731
26,865
27,000
2755
2760
2765
2770
2775
2780
2785
2790
2795
2800
759
762
765
767
770
773
776
778
78i
784
20,911
21,025
21,139
21,254
21,369
21,485
2I,60T
2I,7l8
21,835
21,952
25'55
2560
2565
2570
2575
2580
2585
2590
2595
2600
653
655
658
660
663
666
668
671
673
676
16,679
16,777
16,876
16,975
17,074
17,174
17,274
17,374
17,475
17,576
3005
3010
3015
3020
3025
3030
3035
3040
3045
3050
903
906
909
912
9i5
918
921
924
927
930
27,135
27,271
27,407
27,544
27,681
27,8l8
27,956
28,094
28,233
28,373
2805
2810
2815
2820
2825
2830
2835
28^40
2845
2850
787
790
792
795
798
80 1
804
807
809
812
22,070
22,188
22,307
22,426
22,545
22,665
22,786
22,906
23,028
23,H9
2605
2610
2615
2620
2625
2630
2635
2640
2645
2650
679
68 1
684
686
689
692
694
697
700
702
17,678
17,780
17,882
I7,985
18,088
l8,I9I
18,295
18,400
18,504
18,610
3055
3060
3065
3070
3075
3080
3085
3090
3095
31*00
933
936
939
942
946
949
952
955
$
28,512
28,653
28,793
28,934
29,076
29,218
29,361
29,504
29,647
29,791
2855
2860
2865
2870
2875
2880
2885
2890
2895
2900
815
818
821
824
827
829
832
835
838
841
23,271
23,394
23,5!7
23,640
23,764
23,888
24,013
24,138
24,263
24,389
2655
2660
2665
2670
2675
2680
2685
2690
2695
2700
705
708
710
713
716
718
721
724
726
729
18,715
18,821
18,927
19,034
19,141
19,249
19,357
19,465
19,574
19,683
3105
3110
3115
3120
3125
3130
3135
3140
3145
3150
964
967
970
973
977
980
983
986
989
992
29,935
30,080
30,226
30,37i
30,518
30,664
30,811
30,959
3M07
31,256
2905
2910
2915
2920
2925
844
847
850
853
856
24,515
24,642
24,769
24,897
25,025
Appendix.
443
J
ui
f
I
Cubes.
jt
a
3
i
i
Cubes.

1
ui
z
rt
3
o*
in
Cubes.
3155
3160
3165
3170
3175
2 1 '80
995
999
,002
,005
,008
31,405
31,554
3^705
31,855
32,006
12 I C7
3380
33'85
33'90
33^5
3400
,142
,i 4 6
,149
,153
,156
38,614
38,786
38,958
39.131
39.304
3605
3610
36I5
3620
3625
76*70
1,300
1,303
1,307
1,310
1,314
i 718
46,851
47,046
47,242
47,438
47,635
47 8^2
3185
3190
3195
3200
,014
,018
,021
,024
O^, 1 ^/
32,309
32,462
32,615
32,768
3405
3410
3415
34'20
''59
,163
,166
,170
39,478
39,652
39,826
40,002
O u J^
3635
3640
3645
3650
1,321
1,325
1,329
1,332
liS*
48,030
48,229
48,428
48,627
3205
3210
3215
3220
3225
,027
,030
,034
,037
,040
32,922
33,076
33,231
33,386
33,542
77 608
o4 ^j
3430
3435
3440
3445
3450
, J 73
,176
,180
,183
,187
,190
4 <J , 1 1 1
40,354
40,530
40,708
40,885
41,064
3655
3660
3665
3670
3675
1680
i,336
i,340
i,343
i,347
i,35i
48,827
49,028
49,229
49,431
49,633
AQ 8l6
3 2 3
3235
3240
3245
3250
43
,047
,050
,053
,056
oj> u y
33,855
34,012
34,i7o
34,328
3455
3460
3465
3470
3A"7C
,194
,197
,201
,204
2O8
41,242
41,422
41,602
41,782
3685
3690
3695
3700
,354
1,358
1,362
i,365
1,369
4y,j
50,039
50,243
50,448
50,653
3255
3260
3265
3270
3275
iv 'Xn
,060
,063
,066
,069
,073
34,487
34,646
34,8o6
34,966
35,126
JC 288
34 75
348o
3485
3490
3495
3500
,211
,215
,218
,222
,225
4 i y u j
42,144
42,326
42,509
42,692
42,875
3705
3710
3715
3720
3725
i,373
i,376
1,380
50,859
51,065
51,272
5i,479
51,687
r T 8r>p
3285
3290
3295
3300
,<_>/U
,079
,082
,086
,089
J3, 600
35,449
35,6n
35,774
35,937
3505
3510
35I5
35*20
,229
,232
,2 3 6
,239
43,05 9
43,24 4
43,42 9
43,6i4
At 8no
6/ J U
3735
3740
3745
3750
,39*
i,395
i,399
1,403
1,406
5 r > 5 95
52,104
52,3H
52,524
5 2 ,734
33^5
33!o
33I5
3320
3325
,092
,096
,099
,102
,106
36,101
36,265
36,429
36,594
36,760
35 2 5
3530
3535
35'40
3545
3550
> Z 43
,246
,250
,2 3
,257
260
43,500
43.987
44,174
44,362
44,550
44,739
3755
3760
3765
3770
3775
27*8n
1,410
1,414
1,418
1,421
1,425
52,946
53,157
53,370
53,583
53,796
33 3
3335
3340
3345
3350
,iuy
,112
,116
,119
,122
ju,y^o
37,093
37,260
37,427
37,595
3555
356o
3565
3570
,264
,267
,271
,274
2 ^C
44,928
45, II8
45,308
45,499
67 ou
3785
3790
37*95
3800
1,429
i,433
i,436
1,440
*,444
54,010
54,225
54,440
54,656
54,872
3355
33'6o
3365
3370
33'75
,126
,129
,132
,136
,139
37,764
37,933
38,103
38,273
38,443
DJ /J
358o
3585
35'90
3595
3600
j^7
,282
,285
,289
,292
,296
45,9 r
45,883
46,075
46,268
46,462
46,656
3805
3810
3815
3820
3825
1,448
1,452
i,455
i,459
1,463
55,089
55,306
55,524
55,743
55,962
444
Appendix.
1
Cubes.
Numbers.
I
Cubes.
Numbers.
ri
1
Cube?.
3830
3835
3840
3845
3850
1,467
i,47i
i,475
1,478
1,482
56,l82
56,402
56,623
56,845
57,067
4055
4060
4065
4070
4075
4080
4085
4090
4095
4i"oo
1,644
1,648
1,652
1,656
1,661
1,665
1,669
i,673
1,677
1,681
66,676
66,923
67,171
67,419
67,668
67,917
68,167
68,418
68,669
68,921
42'8o
4285
4290
42'95
4300
1,832
1,836
1,840
1,845
1,849
78,403
78,678
78,954
79,230
79,507
3855
3860
3865
38*70
3875
3880
3885
3890
3895
39*oo
1,486
1,490
1,494
1,498
1,502
1,505
i,59
1,513
1,517
1,521
57,289
57,512
57,736
57,961
58,186
58,4H
58,637
58,864
59,091
59,319
43*05
43IO
43*15
43*20
43*25
43*30
43*35
43*40
43*45
43*5o
1,853
1,858
1,862
1,866
1,871
1,875
1,879
1,884
1,888
1,892
79,785
80,063
80,342
80,622
80,902
8l,l8 3
81,464
81,747
82,029
82,313
41*05
41*10
4115
4120
41*25
41*30
4i*35
4140
4i'45
41*50
1,685
1,689
i,693
1,697
1,702
1,706
1,710
i,7i4
1,718
1,722
69,173
69,427
69,680
69,935
70,189
70,445
70,701
70,958
71,215
71,473
39*05
3910
39*15
39*20
39'25
39*30
39*35
39*40
39*45
39*50
1,525
1,529
i,533
i,537
i,54i
i,544
i,548
1,552
i,556
1,560
59,547
59,776
60,006
60,236
60,467
60,698
60,930
61,163
61,396
61,630
43*55
43*6o
43*65
43*70
43*75
4380
43*85
4390
43*95
4400
1,897
1,901
1,905
1,910
1,914
1,918
1,923
1,927
i,932
i,936
82,597
82,882
83,167
83,453
83,740
84,028
84,316
84,605
84,894
85,184
41*55
4160
41*65
4170
4i*75
4180
41*85
4190
4i*95
4200
1,726
i,73i
i,735
i,739
i,743
i,747
i,75i
i,756
1,760
1,764
71,732
71,991
72,251
72,512
72,773
73,035
73,297
73,560
73,824
74,088
39*55
39*6o
39'65
39*70
39*75
39*80
39*85
39*90
3995
4000
1,564
i!572
1,576
1,580
1,584
I ',592
1,596
1,600
61,864
62,099
62,335
62,571
62,807
63,045
63,283
63,521
63,760
64,000
44*05
4410
44'i5
4420
44*25
44*30
44*35
4440
44*45
44*50
1,940
i,945
1,949
i,954
i,958
1,962
i,967
i,97i
1,976
1,980
85,475
85,766
86,058
86,351
86,644
86,938
87,233
87,528
87,824
88,121
42*05
4210
4215
4220
4225
4230
4235
4240
4245
42*50
1,768
1,772
i,777
i,78i
i,785
1,789
1,794
1,798
1,802
i, 806
74,353
74,6i8
74,885
75,i5i
75,419
75,687
75,956
76,225
76,495
76,766
4005
4010
4015
40*20
4025
4030
4035
4040
4045
4050
1,604
i, 608
i, 612
1,616
1,621
1,624
1,628
1,632
1,636
1,640
64,240
64,481
64,723
64,965
65,208
65,45i
65,695
65,939
66,184
66,430
44*55
4460
44'65
4470
44*75
44'8o
4485
44*90
44*95
45*oo
1,985
1,989
",994
1,998
2,003
2,007
2,012
2,016
2,021
2,025
88,418
88,717
89,015
89,i5
89,615
89,915
90,217
90,519
90,822
91,125
4255
42 '60
4265
4270
4275
1,811
1,815
1,819
1,823
1,828
77,037
77,309
77,58i
77,854
78,128
Appendix.
445
1
3
Cubes.
2
1
I
1
Cubes.
d
1
5
Cubes.
3
fc
fc
1
cl
4505
2,030
91,429
4705
2,214
104,155
4905
2,406
Il8,OIO
45' 10
2,034
91,734
4710
2,218
104,487
4910
2,411
118,371
45I5
2,039
92,039
47'i5
2,223
104,820
4915
2,4l6
"8,733
4520
2,043
92,345
4720
2,228
105,154
4920
2,421
119,095
45 2 5
2,048
92,652
2,233
105,489
4925
2,426
119,459
4530
2,052
92,960
47'3
2,237
105,824
4930
2,430
119,823
4535
2,057
93,268
4735
2,242
106,160
4935
2,435
120,188
4540
2, 06 1
93,577
47*40
2,247
106,496
4940
2,440
120,554
4545
2,066
93,886
47'45
2,252
106,834
4945
2,445
120,920
4550
2,070
94,196
4750
2,256
107,172
4950
2,45
121,287
4555
456o
2,075
2,079
94,507
94,8i9
4755
4760
2,261
2,266
107,511
107,850
4955
4960
2,455
2,460
121,655
122,024
4565
2,084
95,i3i
4765
2,271
108,190
4965
2,465
122,393
4570
2,088
95,444
4770
2,275
108,531
4970
2,470
122,763
4575
2,093
95,758
4775
2,280
108,873
4975
2,475
123,134
4580
2,098
96,072
4780
2,285
109,215
4980
2,480
123,506
2, IO2
96,387
47*85
2,290
109,558
2,485
123,878
4590
2,107
96,703
4790
2,294
109,902
4990
2,490
124,251
4595
2,111
97,019
4795
2,299
110,247
4995
2,495
124,625
46*00
2,116
97,336
4800
2,304
110,592
5000
2,500
125,000
4605
2,121
97,654
4805
2,309
110,938
4610
2,125
97,972
4810
111,285
4615
2,130
98,291
4815
2^318
111,632
4620
2,134
98,611
4820
2,323
III,98o
4625
2,139
98,932
4825
2,328
112,329
4630
2,144
99,253
4830
2,333
112,679
2,I 4 8
99,575
4835
2,338
113,029
4640
2,153
99,897
4840
2,343
"3,380
4645
2,158
100,221
48*45
2,347
"3,732
4650
2,l62
100,545
4850
2,352
114,084
4655
2,l67
100,869
4855
2,357
"4,437
4660
2,172
101,195
4860
2,362
"4.79 1
4665
2,176
101,521
4865
2,367
115,146
4670
2,181
101,848
4870
2,372
46*75
4680
2,186
2,190
102,175
102,503
4875
4880
2,377
2,381
"5^57
116,214
4685
2,195
102,832
4885
2,386
116,572
4690
2,200
103,162
4890
116,930
4695
2,204
103,492
4895
2^396
117,289
4700
2,209
103,823
4900
2,401
117,649
APPENDIX C
SYLLABUS OF EXAMINATIONS IN SUBJECT u.
NAVAL ARCHITECTURE
The students should be encouraged to make good rough
sketches of the different parts of a ship's structure approximately
to scale, using squared paper ; they should also be impressed with
the necessity of noting any detail of work brought before their
notice daily in the shipyard. Questions will be set in the exami
nation which require rough sketches of parts of a vessel to be
given from memory.
If the class is held in an institution which possesses a testing
machine, the students ought to be allowed to use it occasionally to
test samples of materials used in shipbuilding.
All students should be provided with suitable scales, set
squares, and ship curves, and candidates should bring these
to the examination.
Table of logarithms, functions of angles, and useful constants
will be provided, and candidates will be restricted to use of these
tables, and will not be allowed to bring with them into the
examination room any other mathematical or logarithm tables.
Slide rules may be used.
Compulsory questions may be set in either of the examination
papers.
LOWER EXAMINATION
I. PRACTICAL SHIPBUILDING. The tests to which the various
materials used in shipbuilding are subjected, and the defects to
which those materials are liable ; the tools and appliances used in
ordinary shipyard work, and the general arrangement of blocks,
Appendix. 447
staging, derricks, etc., used on a building slip ; plans of flat and
vertical keels, inner bottom, shell, deck and other plating;
framing, beam, keelson, and stringer plans ; watertight and other
bulkheads ; ceiling and wood decks ; pillaring arrangements to
secure clear holds, and details of cargo hatchways to meet Lloyd's
Rules ; rudders, stern frames, and spectacle arrangements for
twinscrew ships ; bilge keels ; supports to engines, boilers, and
shafting ; masts and derricks ; precautions necessary to prevent
deterioration of the hull of a ship while building, and while on
service ; method of docking ships, how they are placed in position
and supported.
II. LAYING OFF. A knowledge of the work carried on in the
Mould Loft for the purpose of fairing a set of lines, including
traces of keelsons and longitudinals, edges of shell plating, tank
margins, ribbands, etc., and transferring the frame and other lines
to the scrive board ; lifting the bevels and constructing round of
beam mould ; a ship's block model and the information necessary
for its construction ; obtaining the dimensions for ordering the
shell plating, frames, beams, floors, inner bottom plating, etc. ;
making and marking ribbands ; fairing the edges of shell plating
on the frames ; making templates or skeleton patterns for stem,
sternpost, propeller bracket forgings or castings.
III. DRAWING. Plotting of curves of displacement, tons per
inch immersion, I.H.P., etc., from given data. A rough freehand
dimensioned sketch may be given at the examination, requiring
candidates to make finished scale drawings, and candidates will
be expected to be able to draw, from their own knowledge, the
fastenings suitable for connecting together the parts which are the
subject of the example.
IV. SHIP CALCULATIONS. Calculation of the weights of
simple parts of a ship's structure ; spacing and strength of iron
and steel rivets ; calculation of the strength of the simple parts
of a ship's structure, such as tie plates, butt straps and laps ; tons
per inch immersion ; change of trim, and moment to change trim;
change of trim due to moving weights on board, and that due to
the addition or removal of weights ; the principles and use of
Simpson's and other rules for finding the area and position of the
centre of gravity of a plane area, and for calculating the position of
the centre of buoyancy ; graphic methods of finding displacement
and position of the centre of buoyancy ; curves of displacement
and of tons per inch immersion ; the fundamental conditions to be
fulfilled in order that any body may float freely and at rest in still
water ; centre of flotation, metacentre, metacentric height, stable
448 Appendix.
and unstable equilibrium ; definitions of block, prismatic, water
plane, midship area, and other similar coefficients.
The questions will be of the same type as those set in Stage 2
of previous examinations.
HIGHER EXAMINATION
I. PRACTICAL SHIPBUILDING. The structural arrangements
necessary to resist longitudinal and transverse stresses to which
ships are liable in still water and amongst waves, and the arrange
ments to resist local stresses ; description and rough hand sketches
of detail fittings of ships, such as anchor and capstan gear, steering
gear, and other appliances used in working a ship ; davits and
fittings in connection therewith ; ventilating and coaling arrange
ments ; pumping and draining ; the fundamental types of vessels
and modifications thereto, the distinctive features of such vessels
and consequent effect on freeboard ; methods of determining the
sizes of structural parts and of detail fittings making out midship
sections to the Rules of the principal classification Societies for
various types of vessels ; methods of fitting up refrigerating spaces
for shipment of frozen and chilled meat, fruit, etc. ; construction of
oil fuel bunkers, and of vessels for carrying oil ; launching arrange
ments, and the diagrams and curves generally used in connection
therewith.
LAYING OFF. Expanding the plating of longitudinals and
margin plates by the geometric and mocking up methods ; ex
panding stern plating, rudder trunking, and mast plating ;
obtaining the true shape of a hawse hole in the deck or shell, and
similar practical problems ; constructing and fairing the form
of a twin screw bossing.
III. SHIP CALCULATIONS. Displacement sheet and arrange
ment of calculations made thereon ; proofs of Simpson's and other
rules for obtaining areas and moments ; displacement and dead
weight scales ; approximate and detailed calculations relating to
the weight and position of the centre of gravity of hull ; calculations
of weight and strength of parts of a ship's structure such as decks,
bulkheads, framing, side and bottom plating, etc., also the strength
of fittings such as boat davits, derricks, etc. ; coefficients of weight
of hull, outfit, and machinery for a few of the principal types of
ships, also coefficients of position of the centre of gravity of the
ships ; curves of loads, shearing forces, and bending moments for
a ship floating in still water, and amongst waves, also equivalent
girder and stress in the material ; calculations of the positions
Appendix. 449
of transverse and longitudinal metacentres ; consideration of the
curves of centres of buoyancy, centres of flotation, and prometa
centres; the construction and use of metacentric diagrams;
Attwood's and Moseley's formulae, and methods of calculating
stability based thereon ; the construction and use of curves of
stability ; inclining experiment and the precautions that must
be taken to ensure accuracy ; change of draught and trim due
to passing from fresh into salt water and vice versa ; effect upon
trim and stability due to flooding compartments of a ship ; effect
of free surface on the stability of vessels carrying liquid cargo ;
methods of determining the size of rudderheads, and the stresses
on rudders balanced and unbalanced ; resistance of ships ;
Froude's experiments on skin friction ; Froude's law of comparison
for vessels at corresponding speeds ; methods of calculating the
horsepower to propel a vessel of known form at a given speed ;
effective horsepower, propulsive coefficient and Admiralty con
stants, and values of the two last in typical cases ; speed of ships
on trial, methods adopted and precautions necessary to obtain
accurate speed data ; progressive trials and their uses ; elementary
considerations of the oscillations of ships in still water and
amongst waves; definitions of a "stiff" and "steady" vessel,
and elements of design affecting these qualities ; tonnage of ships,
how measured, etc.
The questions in this stage will correspond generally in type
with those set in the Stage 3 examination held under the previous
regulations, but the standard will be higher, both for a Pass and
a First Class.
2
APPENDIX D.
1902.
ELEMENTARY STAGE.
General Instructions.
You are permitted to answer only eleven questions.
You must attempt No. II. Three of the remaining questions
should be selected from the Calculations \ and the rest from the
Practical Shipbuilding section.
PRACTICAL SHIPBUILDING.
In Questions I to 9 inclusive your answers may be given in
reference to any type of ship. The type selected should be named
in each question, and the scantlings given in each case.
1. Make a rough sketch showing the crosssection, on a scale
of about ^ the full size, of the flat keel plate and its connections to
the vertical keel plate and to the garboard strake. (6)
2. What is the spacing of the rivets in the edges of the keel
plate and the garboard strake, and in the butt straps of the vertical
keel plate ? (8)
3. Make a sketch showing, for a transverse frame, a cross
section (a) through a floor, and (b) through a frame above the
floor. (8)
4. Make a sketch showing the section of a stem connection to
ordinary outer bottom plating near the waterline. (6)
5. Make a sketch of the crosssection of the arms of an A
bracket or strut of a twinscrew ship, giving dimensions. (6)
6. Make a sketch showing a horizontal section through a
transverse watertight bulkhead. (8)
7. Make a transverse section through the sheerstrake and upper
deck stringer, showing the rivet connection of these plates to the
adjacent strakes and to each other ; also the beam and its rivet
connection to the frames. (8)
Appendix. 45 1
8. Sketch one method of constructing the heads and heels of
pillars, showing the rivet connections. (8)
9. Sketch a butt strap connection of outer bottom plating ;
show the spacing and size of rivets in it. (8)
10. Sketch a disposition of butts in the bottom planking of a
sheathed ship, giving specimens of the positions and sizes of the
fastenings. (10)
DRAWING.
11. Enlarge sketch No. n to a scale of twice that upon which
it is drawn. (This sketch was a portion of the after body of a twin
screw ship, showing frame lines in way of shaft.) (16)
CALCULATIONS.
12. What is the relation which must exist between the weight
of a body floating freely at rest in a liquid and the volume of its
submerged part ? A body of uniform circular transverse section
floats freely in seawater so that the centres of the circular sections
are in the water surface. What will its weight be if its length is
100 feet and its diameter 20 feet ? (8)
13. Find the area of a half of a ship's waterplane of which the
curved form is defined by the following equidistant ordinates
spaced 12 feet apart :
oi, 51, 717, 875, ioi, 917, 805, 64, oi feet. (6)
14. By what number would you have to divide the area in
square feet of a waterplane in order to obtain the number of tons
weight it would be necessary to add to the ship in order to increase
her draught one inch in salt water ? (6)
15. What is the relative position of the centre of gravity of the
weight of a body floating freely at rest in water and the centre of
gravity of the volume of the submerged portion of the body ! What
is the condition necessary for stable equilibrium ? (8)
1 6. What are the weights of a cubic foot of steel, yellow pine,
and copper? What is the weight of a hollow steel pillar 10 feet
long whose external diameter is 5 inches and internal diameter
4 inches ? What is the diameter of a solid pillar of the same
weight? (10)
452 Appendix.
ADVANCED STAGE.
Instructions.
You are permitted to answer only twelve questions.
You must attempt Nos. 22 and 28. The remaining questions
may be selected from any part of the paper in this stage, provided
that one or more be taken from each section, viz. Practical Ship
building, Laying Off, and Calculations.
PRACTICAL SHIPBUILDING.
Questions 17 to 22, inclusive, may be answered with reference
to any one type of ship to which the question may apply, but each
type referred to must be named in the question in which it is dealt
with.
17. Where is the material of a ship's structure most severely
stressed, and under what conditions ? (10)
1 8. Give a sketch showing the disposition and size of rivets in
a buttstrap connection of (i) a sheerstrake, (2) a weatherdeck
stringer bar. Sketch a bulkhead liner. (13)
19. Sketch the transverse framing in a double bottom, giving
the scantlings and the disposition and sizes of the rivets. (14)
20. Describe the operation of framing a ship from the beginning
of handling the unmarked plates and bars to the time the framing
is faired. (20)
21. Sketch the blocks upon which a ship is built, giving the
spacing and sizes of the blocks. (10)
22. Sketch a rudder, giving sizes ; also give the sizes and dis
position of rivets, pintles, and bolts. (13)
23. What is the breaking stress and elongation per cent, in 8
inches of mild steel ? Give the same with reference to any high
tension steel, and name the class of vessel in which it is used.
(10)
Appendix. 453
24. What is the relative elasticity of yellow pine and steel in
combination in a deck in compression and extension ? Sketch an
ordinary disposition of butts and bolts in a wood deck, estimating
the effective area in compression and also in extension. (13)
LAYING OFF.
25. What information is given to the mould loft to enable the
loftman to lay down the lines of a ship ? (10)
26. Show how to obtain the development and the projection in
the sheer and halfbreadth plans of a diagonal. Explain how to
obtain the ending at the stem in the developed diagonal. (18)
27. Show how to find the point where a straight line not parallel
to any of the planes of projection (the sheer, halfbreadth, or body
plan) would cut the surface of the ship. (20)
DRAWING.
28. What does Sketch No. 28 represent ? Enlarge it to twice
the scale upon which it is drawn. (This sketch was the sternpost
and rudder of a screw ship in profile.) (25)
CALCULATIONS.
29. Calculate the volume and position of centre of gravity,
horizontally and vertically, of a form given by the following
ordinates :
ft. ft ft.
No. i W.L. ... o'i 7*17 io  i 8*05 o'i
No. 2 ... o'i 5'66 80 6*46 o'i
No. 3 ... o'i o'i o'i o'i oi
Horizontal interval, 24 feet ; vertical interval, 3 feet.
(18)
30. The areas of transverse vertical sections of a solid are 1*2,
612, 86'o, 1210, 966, 768, 12 square feet, at distances apart of
12, 12, 24, 24, 12, and 12 feet, respectively. Find the volume and
longitudinal position of the centre of gravity of the solid. (10)
454 Appendix.
31. What is the transverse metacentre of a ship in the upright
position ? What is the value of the distance between this meta
centre and the centre of buoyancy ? Find the value of this distance
for the largest waterline given in Question 33 (assuming a value
for the displacement and position of centre of buoyancy if Question
33 has not been attempted). (18)
32. A hold beam is formed of two beams, each formed of a
inch plate 12 inches deep, and four angles 4" x 4" x " I the
beams are connected together by a top plate inch thick, extending
from the fore edge of the flange of the forward beam to the after
edge of the flange of the after beam. Find the weight of 30 feet of
such a beam. The frame spacing is 24 inches. (18)
HONOURS. PART I.
Instructions.
You are premitted to answer only fourteen questions. You
must attempt Nos. 43 and 48 ; the remainder you may select from
any part of the paper in this stage, provided that one or more be
taken from each section, viz. Practical Shipbuilding, Laying Off,
and Calculations.
PRACTICAL SHIPBUILDING.
Questions 33 to 42 may be answered with reference to any
suitable type of ship, but the type must be mentioned, and the
principal figured dimensions inserted.
33 Make a sketch of a sternframe of a twinscrew vessel.
(20)
34. Make a sketch of the method of making the following water
tight in passing through a bulkhead :
(i) A bulb tee ; (2) a keelson formed of a plate and four angle
bars. (13)
Appendix. 455
35. Make a sketch of a boiler hatchcoaming, showing the con
nection to the half ^ earns and to the casingplates. Show what
method is adopted to strengthen the deck at the corners of the
hatch. (15)
36. Make a sketch of a right and lefthanded screw steering
gear, giving the sizes and materials of the different parts. Show
the method of working by hand as well as steam. (21)
37. Sketch the stowage of an anchor, showing the position of
the leads to the capstan and cathead. (21)
38. Sketch the arrangements made for the launch of a ship.
State the declivity and camber of the launching ways, and the
declivity of the keel. (21)
39. Describe the pickling process for steel plates. What pro
tective materials are put upon the various parts of a steel ship ?
(16)
40. Describe briefly the steam pumping and drainage arrange
ments of a ship, detailing where each steampump draws from and
delivers to. (21)
41. Make a sketch showing a section of a steel mast, and show
a disposition of butts of plates, with arrangements of rivets in the
buttstrap. (16)
42. Make a sketch of a pair of davits, giving the cleats and
blocks attached to it. Show the method of securing the boat in
board, and state how the position of the davits is fixed in relation
to the boat. (21)
LAYING OFF.
43. Make a sketch of a body plan on a scriveboard, showing
all the lines that are put upon it, stating what each line is for.
(22)
44. Show how you would obtain the exact, form of the projection
of the intersection of a large conical pipe, with the outer bottom
plating of a ship, the axis of the. cone not being parallel to either
the sheer, halfbreadth, or body plan. (21)
45. Given the body plan of a ship without any bossing out in
way of the shaft, describe how you would obtain the form of this
bossing in the case of a twinscrew ship. (21)
456
Appendix.
CALCULATIONS.
46. Find the displacement up to the 6feet and lofeet water
lines of a ship whose form is defined by the following :
W.L.'s.
Keel.
I ft.
2ft.
4 ft.
6ft.
8ft.
10 ft.
I
Nos. of Sec
Distance
between
sections f
is
40 feet.
tion.
/i
i*
2
3
4
5
6
6*
\7
O'l
O'l
OT
O'l
O'l
O'l
O'l
i '4
2'6
4*6
67
90
II'I
O'l
5'6
8'2
"'5
ISS
H7
I5'3
O'l
ii'i
137
159
167
170
i6'9
O'l
131
I 5 '6
171
171
174
174
O'l
103
I2'6
146
I54
158
160
O'l
57
75
9*5
107
ii'6
125
O'l
17
27
4'i
5'o
60
91
O'l
O'l
O'l
O'l
O'l
O'l
O'l
(21)
47. Find the vertical and longitudinal position of the centre of
buoyancy of the form in the preceding question. (18)
48. Suppose the vessel in Question 54 to be floating at the loft,
waterline, and to be inclined transversely through an angle of one
in one hundred, by a weight of one ton moved through 30 feet.
Find the height of the centre of gravity of the ship above the keel.
(If Questions 46 and 47 have not been done, assume a displace
ment and height of C.B.) (18)
49. What is the ultimate shearing and tensile stresses of steel
rivets and plates respectively? Find the breaking strength of a
single butt, doublechain riveted, of a plate 30 inches wide, inch
thick, connected by rivets, spaced 3 inches apart. Find the force
necessary to break the plate across a frameline where the rivets
are spaced 6 inches apart. (21)
50. Given the height of the longitudinal metacentre above the
centre of gravity, show how you would obtain the moment to trim
ship one inch. In a vessel whose distance between draft marks
forward and aft is 300 feet, and whose centre of gravity of water
plane is 10 feet abaft centre between draft marks, and whose foot
Appendix. 457
tons to trim ship one inch are 400, find the change of draft forward
and aft caused by moving 30 tons through 200 feet in a fore and
aft direction. (21)
51. A barge 100' x 20' x 10' of rectangular section, is formed
of inch plating on ends, bottom, sides, and deck, and has frames
and beams of 4^" x 4" x ", spaced 20 inches apart. The ends
have stiffeners 2 feet apart, of the scantlings of the frames. A floor
plate, 12" x ", is on every frame. Find the weight of the hull,
assuming that there are no hatches. Suppose the barge to have
weights of 10 tons at 10 feet from the stem, 15 tons at 25 feet, 20
tons at 50 feet, 30 tons at 75 feet. Find the longitudinal position
of the centre of gravity of the loaded barge. (30)
HONOURS. PART II.
Instructions.
You are not permitted to answer more ihanfvurteen questions.
NOTE. No Candidate will be credited with a success in Part
II. of Honours who has not obtained a previous success in Honours
of the same subject.
Those students who answer the present paper sufficiently well
to give them a reasonable chance of being classed in Honours,
will be required to take a practical examination at South Kensing
ton. Honours Candidates admissible to this Examination will be
so informed in due course.
52. State and prove Simpson's First Rule. State Tchebycheffs
Rules for either three, five, or seven ordinates. (25)
53. Prove that in a curve of loads of a ship floating at rest, the
integral of the area of the curve from one end up to a chosen point
gives the shearing force at that point, and that the integral of the
curve of shearing forces over the same part gives the bending at
the chosen point. (25)
54. Suppose a curve of buoyancy to be a curve of versed sines,
and the corresponding curve of weights to be a common parabola,
whose axis is vertical and at the middle of the length. Find the
form of the curves of shearing force and bending moment. (33)
55. State and prove Atwood's formula. (21)
56. Describe any method of obtaining a crosscurve of stability.
(25)
45 8 Appendix.
57. Find the effect upon the draught of water forward and aft
of opening to the sea one per cent, of the length of a vessel of
rectangular section at any part of the length, supposing this one
per cent, to be confined between transverse water tight bulkheads.
Explain how the deductions from this result can be made use of
to determine the spacing between water tight bulkheads, which
shall not be exceeded, in order that when a compartment is flooded,
the draft in no case shall exceed a certain specified amount. (33)
,58. If B and B l be respectively CB's in upright and inclined
position, and R be the foot of the perpendicular from B on to the
vertical through B^ in this inclined position, show that B\R is the
integral of BR between the upright and the angle of inclination.
Show from this how a curve of CB's can be obtained from a curve
of GZ's. (25)
59. A vessel of uniform rectangular section is launched parallel
to her keel. What is the form of the curve of tipping and lifting
moments, supposing the ship to be deep enough to prevent the
upper deck from being immersed ? Suppose such a vessel 100 feet
long, 24 feet wide, having its CG at the middle of the length to be
launched at a slope of one inch to the foot, with two feet of salt
water over the end of the ways, and with a launching weight of 100
tons. What is the maximum pressure on the fore end of the ways
(assumed to be at the fore perpendicular) ? (43)
60. Suppose a vessel to be instantaneously floating on the crest
of a wave of her own length, and of height equal to onetwentieth of
the length. What stress would you expect to find with all coal
burnt out in
(1) a battleship of 14,000 tons ;
(2) a highspeed Atlantic liner ;
(3) a torpedoboat destroyer ?
Give figures for the stress when in the hollow of the same wave with
bunkers full. (25)
61. Why would a trochoidal wave cause less stress than that
determined in the preceding question? Give results of any
calculations you know of, in which the difference due to the wave
not being actually at rest is taken into account. (25)
62. Prove that in a ship, whose moment of inertia about a
transverse axis through the midship section is the same for the
fore end as for the after end, the pitching does not alter the bend
ing moment at the midship section. (25)
63. Find the maximum stress upon a section of a vessel floating
upright in still water, and subjected to a bending moment of 1000
Appendix.
foottons. The section is rectangular, 20 feet wide, 10 feet deep,
and has " plating on deck, bottom, and sides.
Suppose the vessel to be inclined at some known angle, how
would you find the maximum stress ? (33)
64. What is the Admiralty speed coefficient ? What is its value
for the vessels named in Question 60 at full speed? How does it
vary with speed? What is its value for sea work as compared
with trial trips ? (33)
65. A model 12 feet long, having a displacement of 1000 Ibs.,
has a resistance of 3 Ibs. at 5 feet per second. Find the effective
horsepower necessary to drive a vessel of the same form 192 feet
long at its corresponding speed. Assume the wetted surface of the
model to be 30 square feet, and the frictional resistance of a plane
12 feet long at 5 feet per second to be 0*07 Ib. per square foot, and
that of a plane 192 feet long to be o'8 Ib. per square foot at 12
knots. (25)
66. What conditions should be fulfilled in a ship to make her
easy in her rolling at sea ? (25)
67. What is a curve of extinction? How can it be obtained
experimentally? What can be determined from it in relation to
the resistance to rolling of a ship ? (25)
68. What is the chief cause of vibration in a steamer ? What
are the subsidiary causes ? What precautions are taken to avoid
these ? (25)
1905.
STAGE II.
Instructions.
You are permitted to answer only eight questions.
You must attempt Nos. 32 and 35. The remaining questions
may be selected from any part of the paper in this stage, provided
that one or more be taken from each section, viz. Practical Ship
building, Laying Off, and Calculations.
PRACTICAL SHIPBUILDING.
21. Sketch a good disposition of butts for outer bottom plating,
assuming your own spacing of frames and length of bottom plates,
460 Appendix.
stating the spacing of frames and length of plates you have
assumed. (20)
22. For what purposes are " liners " and " bulkhead liners "
fitted in ships ? Describe a system of construction in which
ordinary liners are not necessary. Sketch a bulkhead liner. (20)
23. Sketch and briefly describe any efficient type of steering
gear. (20)
24. State what parts of a vessel are most effective in resisting
longitudinal stresses, and give the reasons for your answer.
(25)
25. Sketch and briefly describe the construction of a steel mast,
stating the size of mast, scantlings, and size and pitch of rivets.
(20)
26. Show by sketches how a modern hawse pipe is secured to
the structure of a ship. Of what materials are hawse pipes made,
and how is the diameter determined ? (20)
27. How is a large transverse watertight bulkhead plated, stiff
ened, and secured ? (25)
28. State the tests made to ensure that either large or small
steel castings, and steel plates are fit for use in shipwork. Describe
how the millscale formed on steel plates during manufacture can
be removed. (20)
LAYING OFF.
29. Briefly describe the contracted method of fairing the body.
If the floor be of small length, what precautions are necessary when
fairing a long ship ? (20)
30. How would you obtain the lines of the inner bottom on the
floor, and fair them ? How would you arrange and fair the plate
edges ? (20)
31. Describe how an account is obtained of the outside plating
of a steel ship in order that the plates may be demanded from the
manufacturer. What margin would you allow at the edges and
butts ? (25)
DRAWING.
32. What does the given sketch represent? It is drawn on a
scale of " to i foot ; draw it in pencil on a scale of " to I foot.
(35)
CALCULATIONS.
33. The tons per inch immersion of a ship at seven equidistant
waterlines 3 feet apart are respectively 313, 302, 285, 264, 23 9,
196, and 142. Find the displacement and the vertical position of
Appendix, 4^ T
the centre of buoyancy. The appendage below the lowest water
plane to be neglected. (22)
34. What conditions have to be fulfilled in order that any body
may float freely and at rest in still water ? What is the condition
necessary for stable equilibrium ? (20)
35. State Simpson's second rule.
The equidistant halfordinates of the load waterplane of a ship
in feet are o'6, 2*9, 91, 15*6, i8'o, 187, 185, 176, 152, and 67
respectively, and the length of the ship is 288 feet. Find the area
of the load waterplane and the longitudinal position of the centre
of gravity. (25)
36. Describe the process known as the "graphical process,"
used for finding the displacement and centre of buoyancy of a ship.
(23)
STAGE III.
Instructions.
You are permitted to answer only eight questions. You must
attempt No. 51, and one other question at least should be selected
from the Calculations.
PRACTICAL SHIPBUILDING.
41. To what stresses is the hull structure of a ship subjected ?
Describe how they are set up, and the provision made to meet
them. (35)
42. Describe briefly the principal pumping and draining arrange
ments of a large ship, naming the type of ship selected. (35)
43. Roughly sketch the midship section of a vessel, giving
scantlings. Name the type of vessel selected. (35)
44. Sketch and describe the usual methods of constructing and
fitting watertight sliding doors. How are they opened and closed ?
(35)
45. Describe, with sketches, the construction and uses of the
following shipyard machines, viz. :
(a) Plate bending rolls.
(b} Punching and shearing machine.
Show in detail the construction of the latter machine in way oi
the punch. (35)
46. Sketch and describe the special features of the general
arrangement and detail construction of the hull of an oil steamer.
(35)
462
Appendix.
47. Sketch and describe the construction of a bridge. How is
it supported and stiffened against rolling strains ? Enumerate the
fittings generally placed on the bridge, and show how they are
arranged. (35)
48. Describe how the scantlings of a ship of known type and
dimensions are determined by Lloyd's Rules. What portions of
the hull structure are determined by the ist and 2nd numbers
respectively ? (40)
LAYING OFF.
49. Describe how the line of centre of shaft is got in on the
floor, and the body faired into the shaft tube of a twin screw ship.
(35)
50. How would you obtain a correct mould for a longitudinal,
or tank margin plate, which has considerable twist and curvature ?
What marks would be placed on the mould for the information
and guidance of the workman ? (35)
CALCULATIONS.
51. Calculate the displacement and vertical position of the
centre of buoyancy of a vessel for which the halfordinates are
given below, the distance between the sections being 14 feet,
and the keel appendage being 2'6 tons, with centre of buoyancy
48 feet below the 5' 6" waterline.
(45)
Sections.
,W.L.
i' 9" W.L.
a' 6" W.L.
4 ' W.L.
5 '6"W.L.
!
O'l
o3
08
2'8
2
2'2
39
5'4
66
7*
3
3'4
6'8
7*4
7*5
4
2*O
3*6
5' 1
64
68
5
O'l
0'2
1*4
28
52. Define change of trim) and moment to change trim one inch.
Obtain an expression for the position in which a weight must
be placed on board a ship so as not to increase her maximum
draught. Explain clearly why this is not always possible with
large weights, and find the limiting weight. (35)
53. What are the curves of displacement and tons per inch
immersion, and what are their uses ?
Appendix. 463
The areas of a ship's sections at parallel waterlines 3 feet apart
are 9600, 9500, 9000, 7700, 5000, and 2000 square feet. Neglecting
the volume below the lowest section, find the tons per inch at each
waterplane, and plot the curve of tons per inch. Find also the
total displacement. (35)
54. What are the ultimate shearing and tensile stresses of steel
rivets and plates respectively ?
Two tie plates 24" wide by f" thick, are connected together by
a lapped joint. Show by calculation the number and sizes of rivets
required, indicating how they should be arranged in order that the
butt and plate may be nearly of equal strength. (35)
55. Define "centre of flotation," " centre of buoyancy," "meta
centre," and " metacentric height."
Determine the distance between the centre of buoyancy and
the transverse metacentre of a vessel 72 feet long and 95 tons
displacement, floating at a waterplane whose halfordinates are,
08, 3'3, 5*4, 65, 68, 63, 51, 28, and 05. (40)
56. A foreand aft watertight bulkhead, extending from the
tank top to main deck, is 50 feet long and 24 feet deep. Find the
total weight of the bulkhead, including stiffeners, connecting angles,
etc., having given the following particulars :
Plating jV thick for the lower half depth, and J" above, with
singleriveted edges and butts ; stiffeners alternately 6" x 3" x 3^"
zed bars of 15 Ibs. per foot run, and 3^" x 2^" angle bars of 7 Ibs.
per foot run, spaced 2 feet apart ; bounding angles 3^" x 3" of
8'5 Ibs. per foot run. (35)
HONOURS.
Instructions.
You are not permitted to answer more than eight questions.
NOTE. No Candidate will be credited with a success in this
examination who has not obtained a previous success in Stage II 7.
or in Honours, of the same subject.
61. State and prove Simpson's second rule for approximating
to the area and centre of gravity of a plane surface.
The halfordinates of a waterplane are 0*2, i'8, 4'8, 7*4, 5*5, 2*3,
and 0*6 feet. The ordinates are spaced 23 feet apart. Find the
distance of the centre of gravity of the half waterplane from the
middle line. (40)
464 Appendix.
62. Obtain an expression for the height of the metacentre above
the centre of buoyancy in a floating body.
The half ordinates of the waterplane of a vessel, 27^ feet apart
are o'i, 6*9, io'o, 105, 10*1, 7*2, and o'i feet respectively. Deter
mine the transverse metacentric height, having given that the dis
placement to the water plane (salt water) is 275 tons, and that the
centre of gravity of the vessel is 5 feet above the centre of buoyancy.
State the values of the metacentric heights in any two types
of ships with which you are acquainted, naming the types of vessels
selected. (50)
63. The maximum speed of a vessel is 17 knots, and the rudder,
which is 12 feet broad and approximately rectangular in shape, has
an area of 200 square feet and a maximum working angle of
35 degrees. Estimate the diameter required for the rudder head
if made of cast steel.
How does the case of a balanced rudder differ from that of an
ordinary one in the case (a) when the ship is going ahead, and (b}
when she is going astern ? (45)
64. What are " cross curves of stability " ?
Describe fully how you would construct a set of cross curves of
stability for a vessel of known form. Explain clearly the great
advantages of having stability calculations recorded in this form.
(45)
65. Show how you would estimate the angle of heel to which
a ship under sail in still water would be driven, when struck by a
squall of known force, (a) when the ship is upright and at rest ; and
() when the ship has just completed a roll to windward, when the
squall strikes her. (45)
66. Prove that in a curve of loads of a ship floating at rest, the
integral of the area of the curve from one end up to a chosen point
gives the shearing force at that point.
At the section of a ship at which the shearing force is at a
maximum, show how the shearing stress on the material varies,
and state under what circumstances this shearing stress would cause
straining action to take place. (50)
67. Describe the principles governing the watertight subdivision
of war or merchant ships. Is there any legal enforcement for
merchant ships ?
State briefly the recommendations of the Bulkheads Committee
(189091).
A barge is of uniform rectangular section, 70 feet long and
20 feet broad, and the draught of water when the vessel is
intact is 8 feet. What would be the minimum height of a bulkhead
Appendix. 465
lo feet from one end of the vessel in order that if the end compart
ment were flooded, the adjacent compartment should remain dry ?
(50)
68. Define Statical Stability and Dynamical Stability.
A submarine vessel 140 feet long has a uniform cross section
of which the upper part is a semicircle 10 feet in diameter, and the
lower a triangle 8 feet deep with vertex downwards. The centre of
gravity of the vessel is 6 feet above the keel.
Construct, to scale, the curve of statical stability, and state
in foottons the dynamical stability at 60 degrees. (50)
69. State what is meant by " effective horsepower," " propulsive
coefficient," and " corresponding speeds." State the values of the
propulsive coefficients of any two types of vessels with which you
are acquainted, naming the types selected. How does the pro
propulsive coefficient vary with the speed in a particular ship, and
why?
A vessel of 1800 tons displacement is propelled at 15 knots by
engines of 2500 I.H.P. Estimate the I.H.P. you would consider
necessary to drive a vessel of similar model, but of 4000 tons dis
placement at a speed of 18 knots. What assumptions are made in
passing from the one vessel to the other ? (45)
70. Describe briefly the causes which produce vibration in the
hulls of steamships, and state under what circumstances these
vibrations reach a maximum.
(a) State whether you consider vibration to be indicative of
structural weakness, giving reasons for your answer.
(b] How would you attempt to reduce vibration when excessive ?
(f) At the lowest number of vibrations possible, where would
you expect to find the nodal points ?
(d) What recent modifications in design are known to produce
less vibration ? (50)
71. Describe fully the method of conducting measuredmile
trials and arriving at the measuredmile speed.
State the possible sources of error to which such trials are
liable, and how they are reduced. (40)
72. How would you obtain the wetted surface of a ship of known
form ?
Quote any formula giving a close approximation to the wetted
surface.
What use can be made of the wetted surface when obtained ?
The wetted surface of a ship of 6000 tons displacement being
25,000 square feet, find the wetted surface of a vessel of similar
form, but of 2000 tons displacement. (40)
2 H
466 Appendix.
73. A vessel runs bowon to a shelving beach ; investigate her
stability, as compared with her condition when afloat.
A boxshaped vessel, 100 feet long and 20 feet broad, floats at a
draught of 6 feet forward and 10 feet aft, the metacentric height
being T.\ feet. Find the virtual metacentric height when she just
grounds all along on level blocks. (45)
74. Sketch and describe the launching arrangements for a large
ship, stating the dimensions of the vessel, declivity of the blocks
and launching ways, and the pressure per square foot allowed on
the surface of the ways. What is the meaning of the term
" camber " as applied to the ground ways, and to what extent is it
admissible ?
The launching weight of a ship is 2800 tons, its centre of gravity
is 9 feet abaft the midship section, and the fore end of the launching
cradle is 120 feet before the midship section. When the midship
section of the vessel is respectively o, 10, 20, 30, 40, and 50 feet
abaft the after end of the ways, the corresponding buoyancy is
respectively mo, 1310, 1530, 1770, 2030, and 2310 tons, and the
distances of the corresponding centres of buoyancy abaft the after
end of the ways are respectively 43, 51, 60, 68, 77^, and 86 feet.
Construct to scale the corresponding launching diagrams,
stating where the stern begins to lift, and the pressure on the fore
poppet. Are the ways sufficiently long to prevent tipping ? (55)
75. Investigate the value of the metacentric height of a vessel
with free water in the hold.
A mud hopper of box form is 200 feet long and 40 feet broad,
the mud chamber being the amidships portion 50 feet long. When
empty, the draught is 10 feet and the centre of gravity 15 feet
above the keel. Find the metacentric heights when (a) Empty,
() Dischargeport is open, and (c) Chamber is filled to a height of
10 feet with sludge of specific gravity 2. (45)
76. Describe in detail how you would proceed to fix the dimen
sions and underwater form of a combined passenger and cargo
carrying steamer, having given the speed, length of voyage,
maximum draught permissible, cargo capacity (both by measure
ment and dead weight), number of passengers, and type of vessel.
State what you consider satisfactory limits of stability for such
a vessel as you select. (50)
Appendix. 467
1908.
STAGE II.
Instructions.
You are permitted to answer only eight questions.
You must attempt Nos. 32 and 33 ; also three questions in
the Practical Shipbuilding Section, and one in the Laying Off
Section. The two remaining questions may be selected from any
part of the paper in this stage.
PRACTICAL SHIPBUILDING.
21. Describe, and show by sketches in section and side
elevation, how an intercostal plate keelson (or vertical keel) is
worked and secured in an ordinary transversely framed vessel
with a flatplate keel. (20)
22. Show how a large transverse watertight bulkhead is built,
stiffened, and riveted.
How would you check the position of such a bulkhead at
the ship, and how would you test its watertightness when
completed ? (25)
23. Sketch the portion of the midship section extending from
the margin plate to the upper deck in a mercantile vessel, or,
from the 4th longitudinal to the upper deck in a war vessel. Show
the arrangement in detail, with scantlings of outer bottom plating,
framing, stringers, etc., and name the type of vessel selected.
(22)
24. Sketch and describe the construction of the stem of any
large vessel, showing in detail its connections to the keel, decks,
and shell plating. Name the type of vessel selected, and state
the material of which the stem is made.
What tests would you apply to ascertain the fitness of the
stem ? (20)
25. Sketch a good shift of butts for the tank top or inner
bottom plating of a vessel, with reference to the butts of the
girders or longitudinals. Give details of the butt and edge con
nections of the plating, and state the spacing of the frames, length
of plates, etc. (22)
26. Sketch in detail, and describe a good arrangement of
boilerbearers, giving scantlings, etc., and name the type of vessel
selected. (20)
468 Appendix.
27. Compare the qualities of " mild " and " hightensile " steel,
and state the tests you would expect hightensile steel plates to
satisfactorily comply with.
In what parts of the structure of a ship would it be of advantage
to fit hightensile steel, and what precautions should be taken in
working it ? (23)
28. What permanent arrangements are made for protecting the
side of a ship which has frequently to lie alongside quay walls, etc. ;
such as cargo vessels, tugs, etc.
Sketch and describe the construction of any suitable arrange
ment, giving scantlings and particulars of the fastenings, and state
where such an arrangement should be fitted. (20)
LAYING OFF.
29. Define the following terms, viz. " Length over all,"
" Length between perpendiculars," " Moulded breadth," " Moulded
depth," " Rise of floor," " Dead flat," " Tumble home," " Depth of
hold," " Camber of beam," and " Bilge diagonal." (20)
30. How would you make a beam mould ? Show how to obtain
the beam end line in the sheer and body plans, taking into account
the roundup and sheer of the deck. (22)
31. What information would be required, and how would you
proceed to demand the materials, for rapidly building a steel
vessel, indicating the order in which you consider the materials
should be demanded ?
What margins are allowed over the actual dimensions
required? (25)
DRAWING.
32. The given sketch represents not to scale five equidistant
sections, obtained by taking measurements from the outside of
a vessel in dry dock. Offsets to the various water lines, level
lines, and bow lines are given in the table accompanying the
drawing.
With the given offsets and particulars on the drawing, draw
the five sections neatly, in pencil, to a scale of \" equals i foot.
(35)
CALCULATIONS.
33. Write down and briefly explain the rules in common use
in ship calculations, for finding the areas of plane surfaces and
volumes of displacement.
The semiordinates of the boundary of a deck of a vessel
Appendix. 469
are :~o*5, 4*5, 88, io'o, 8'2, 38, and 04 feet respectively, including
the end ordinates. The length of the vessel being 85', find the
area of the deck, and the position of its centre of gravity. (25)
34. A deck of a vessel is composed of flush plating &" thick,
secured to channel bar beams 8" x 4" x ", spaced 3' 6" apart.
Calculate the weight of a part of the deck 63' long by 10' wide,
including rivets, but omitting edge strips and butt straps. (20)
35. What are the curves of " displacement " and " tons per inch
immersion," and what are their uses ?
The area of a ship's load water plane is 6050 square feet, the
body below is divided by equidistant horizontal sections 3' apart,
whose areas are 5500 ; 4750 ; 3500 ; 2050 ; 1000 ; and 250 square
feet respectively.
Find the tons per inch at each water plane, and plot the curve
of tons per inch, on the squared paper supplied.
What is the total displacement of the vessel ? (23)
36. Explain, in detail, a method of determining graphically the
displacement of a vessel of given form. (22)
STAGE III.
Instructions.
You are permitted to answer only eight questions.
You must attempt No. 52. The remaining questions may be
selected from any part of the paper in this stage, provided that one
or more be taken from each section, viz. Practical Shipbuilding, Lay
ing Off, and Calculations ; but you must not attempt more than
three questions ', including No. ^from the Calculations section.
PRACTICAL SHIPBUILDING.
41^ Describe with the aid of rough sketches, the fittings neces
sary for efficiently working the anchors of any vessel with which you
are acquainted, naming the type of vessel selected. (35)
42. Describe the difference between an " ordinary " and a
"balanced" rudder, and tate why the latter are sometimes fitted
in ships.
Sketch a sternpost suitable for a balanced rudder, and show how
it is connected to the keel, decks, and shellplating. (35)
43. Roughly sketch the midship section of any vessel with which
you are acquainted, naming the type of vessel selected.
Indicate the scantlings of the various parts comprising the
section. (35)
4/0 Appendix.
44. Sketch and describe in detail the construction of an accom
modation or gangway ladder, showing how it is raised and stowed.
(35)
45. What precautions are taken in oilcarrying steamers to avoid
risk Of explosion ?
How are the bulkheads of such vessels constructed? Give
scantlings of plates, etc., and size and spacing of rivets. (35)
46. Describe, with sketches, the construction of a horizontal
punching machine, and explain how the machine can be adapted
to do riveting, beambending, and angle bar cutting. (35)
47. Explain how the Rules of Lloyd's Register determine the
scantlings of a threedecked merchant steamer, distinguishing
between the transverse and the longitudinal portions of the
structure.
In a vessel as above, it is desired for convenience of stowage
to omit the " hold " beams. Describe the modifications you would
adopt to the ordinary construction consequent upon that omission.
(40)
48. Describe, with sketches, the method of coaling a large ship,
such as an Atlantic liner or a warship, from a collier or barge
alongside.
Show, by a sectional sketch of the vessel, how the coal is passed
from the upper or coaling deck, to the bunkers. (35)
49. Sketch and describe the construction of a steel deckhouse.
In the case of a vessel which has a long bridgehouse to be
worked amidships, how should this bridgehouse be constructed in
order that it maybe made an efficient pa/t of the structural strength
of the vessel? (35)
LAYING OFF.
50. Distinguish between a " ribband," a " harpin," and a " sheei
harpin."
Show how to lay off and obtain the bevellings of a sheer
harpin. (33)
51. The lines of a steel vessel, sheathed with wood, having been
given to the outside of sheathing, show how you would obtain Ithe
body plan to outside of framing (i) approximately, and (2)
accurately. (35)
CALCULATIONS.
52. The given sketch represents part of the fore body of a ship.
Calculate the displacement in tons, and the vertical position of the
centre of buoyancy of the form represented by the sketch, between
Appendix. 47 j
the waterlines A and B, 12' apart, and between the sections C and
D. The sketch given is on a scale of \" equals I foot, and the
sections are spaced 20' apart. Four waterlines at depths of 3', 6',
9', and 10' 6" below A waterline are to be introduced between A
and B for the purpose of the calculations.
Ordinates to be measured to the nearest first decimal place.
(45)
53. State and prove Simpson's ist Rule, for approximating to
the area and centre of gravity of a plane surface.
The equidistant halfordinates of a waterplane being 3*0, 5*4,
7 1, 932, 122, 1417, and 195 feet respectively, and the length of
the base being 84*0 feet, find the area of the waterplane, and the
transverse position of the centre of gravity of half the waterplane.
(40)
54. The halfordinates of a portion of a ship's deck, covered with
&" plating, are 42, 936, 123, 1484, 165, 1753, and 187 feet in
length respectively, the common interval being 1 5 feet.
Calculate the weight of the beams, plating, planking, and
fastenings, etc., for this part of the deck, the beams being 8" X 4"
x j$y" channel bar, spaced 3' 6" apart, and the plank being of pitch
pine 3$" thick.
Estimate the cost of laying the deck with planks 6" wide at
1\d. per foot run. (35)
55. What is the ultimate shearing and tensile stresses of mild
steel rivets and plates respectively ?
The shell plating of a vessel is formed of plates 50" wide and
^" thick, worked on the raised and sunken system ; the spacing
of the rivets in the frames are 7 diameters apart, and in the
boundary angles of the watertight bulkheads 4 diameters apart.
Sketch an arrangement you would make in order that the strength
of the shell plating in wake of the bulkheads and ordinary frames
shall be approximately the same. Show, by calculations, that
your arrangement is a good one. (35)
56. Define " centre of gravity," " centre of buoyancy," " centre
of flotation," " metacentre."
A vessel 140' long, and whose body plan halfsections are
squares, floats with its sides upright, and the centres of all the
sections lie in the plane of flotation. The lengths of the sides of
the sections, including the end ordinates, are o'8, 3 '6, 7'o, 8'o, 6*4,
3'o and 07 feet respectively, equispaced.
Calculate the distance between the centre of buoyancy and the
transverse metacentre. (35)
472 Appendix.
HONOURS.
Instructions.
You are not permitted to answer more than eight questions.
Note. No candidate will be credited with a success in this
examination who has not obtained a previous success in Stage III.
or in Honours, of the same subject.
Those candidates who do well in the following paper will be
admitted to a practical examination held at South Kensington or
some other centre. Candidates admissible to that examination will
be so informed in due course. No candidate will be classed in
Honours who is not successful in the Practical Examination.
61. Having given the value of six equidistant ordinates of a plane
curve, deduce a formula that will give the area of the surface lying
between the extreme ordinates and the curve.
Four consecutive polar radii of a curve, taken in order, are
io'9, 1 1 '6, 13x3, and 14/1 feet; the common angular interval
between them is 15 degrees. Find the area, in square feet, included
between the curve and the extreme polar radii, and prove the rule
you use. (45)
62. Obtain an expression giving the height of the longitudinal
metacentre above the centre of buoyancy. What use is the
information when obtained for any particular vessel ?
Draw, to scale, the ordinary metacentric diagram for a vessel
whose uniform section throughout her length is a quadrilateral of
breadth 50' at the load line and 25' at the keel, the draught of
water being 20'. (45)
63. Under what circumstances may it be expected that the
cargoes of vessels will shift ?
In a cargocarrying vessel, the position of whose centre of
gravity is known, show how the new position of the centre of
gravity, due to a portion of the cargo shifting, may be found.
A ship of 4800 tons displacement, when fully laden with coals,
has a metacentric height of 2*6 feet. Suppose 120 tons of coal to
be shifted so that its centre of gravity moves 19 feet transversely
and 5 feet vertically, what would be the angle of heel of the vessel,
if she were upright before the coal shifted ? (45)
64. Prove that for any floating body revolving about an axis
fixed in direction, positions of maximum and minimum stability
occur alternately.
Investigate all the positions of equilibrium for a square prism
Appendix. 473
of uniform density revolving about a horizontal axis, assuming its
density to be threefourths that of the fluid it is floating in. (50)
65. Quote Moseley's formula for the dynamical stability of a
floating body, and prove that the value of the dynamical stability
obtained from that formula is identical with that obtained by
integrating the curve of statical stability.
A vessel of constant rectangular section is 260' long, 30' broad,
30' deep, and draught of water 15'. The metacentric height of the
vessel being 25', find (i) the statical stability, 0</(2)the dynamical
stability of the vessel when she is inclined at 45 degrees. (50)
66. A boxshaped vessel 420' long, 72' broad, and draught of
water 24', has a compartment amidships 60' long, with a water
tight middle line bulkhead extending the whole depth of the vessel.
Determine the angle of heel caused by the ship being bilged on
one side abreast this bulkhead, the centre of gravity of the vessel
being 23' above the keel.
To what height should the transverse bulkheads at the ends of
the bilged compartment be carried, so as to confine the water to
this part of the vessel ?
If the bilging be caused by a collision, making a hole 1*5 square
feet in area at a depth of 18' below the load waterline of the
vessel, in wake of the compartment referred to above, and the
pumps be in working order, calculate the capacity of the pumps
required to just keep the leak under. (45)
67. Define " freeboard," and state what determines it. Describe
the arrangement of the tables giving " freeboard."
How is the statutory deckline marked ?
Distinguish between "flushdeck," " spardeck," and " awning
deck" vessels. How is the freeboard determined in each case ? (50)
68. Prove the relation which exists between the load curve and
the curves of shearing force and bending moment.
A vessel 300' long has a uniform section below water. The
weights of hull, machinery, and cargo are 840, 300, and 300 tons
respectively. The weight of machinery extends uniformly over
rd of the length amidships, and the weight of cargo extends
uniformly over th of the length from each end. The weight of
hull curve is of the form
I
1H
I
>k
FIG. 157
t
474 Appendix.
Draw the curves of shearing force and bending moment and
state their maximum values when the vessel is at rest in still
water. ( 45 )
69. The effective part of the transverse section of a vessel amid
I ships is represented by the dia
gram (Fig. 1 58), the vessel being
<o I 42' broad and 28' deep.
Find the maximum tensile
and compressive stresses when
the vessel is subjected to a sag
ging bending moment of 60,000
foottons. The plating shown in
the diagram to be taken as i"
u 14.' >{ 14.' I J< 4/ J thick, and no allowance need be
' made for rivet holes or laps of
FlG  '5 8 ' plating.
Assuming I* and I v are the moments of inertia of a section
about axes at right angles to each other, deduce a formula for
finding the stress on the section at any point when a vessel is
inclined at an angle to one of the axes. (50)
70. Enumerate the component parts of the total resistance to
propulsion of a ship. What is the relative importance of these
component parts at (i) low speeds, and (2) at high speeds?
A ship, 290' long, 45' beam, 17' 6" draught, and 3200 tons
displacement, steams 17^ knots. Find the horsepower necessary
to overcome frictional resistance, having given that the resistance
varies as the 183 power of the velocity, and that in fresh water, at
a speed of 10 feet per second, the average resistance for a length
of 50' is 0*246 Ibs. per square foot, whilst over the last square foot
the resistance is 0*232 Ibs. (45)
71. Explain in detail how the indicated horsepower for a new
ship is estimated.
A model of a vessel, 400' x 65' x 24' draught, of 8560 tons
displacement, is run, and the curve of E.H.P. on a base of speed
of ship is 3250, 4035, 5020, 6195, and 7660 E.H.P. for 16,
17, 18, 19, and 20 knots respectively. Make an estimate of the
I.H.P. of a ship of 16,000 tons, of similar form, for speeds of 20 and
21 knots, and give the dimensions of the new ship. (45)
72. The draught of water, the desired speed, and the load to be
carried being given for a new design, state in detail how you would
obtain the approximate dimensions of the ship.
Obtain suitable dimensions for a vessel to carry uoo tons of
cargo on a limiting draught of 21', the speed of the vessel to be
Appendix. 475
12 knots, with coal sufficient for a voyage of 1500 miles, and
300 tons of passengers and stores. (50)
73. Deduce a formula for the period of a ship whose rolling is
unresisted and isochronous.
A vessel of 13,500 tons displacement has a metacentric height
of 3*5 feet and a period of 8*5 seconds. Find the period of rolling
when 600 tons of coal are added each side of the vessel in a bunker
21' deep and 9' wide, the centre of gravity of the bunkers being
n' below the original centre of gravity of the ship and 26' out
from the middle line. The vessel has a horizontal curve of
metacentres over the limits of draught corresponding to the above
conditions. (50)
74. Define the terms "effective wave slope" and "virtual
upright." Explain under what circumstances the rolling of a
vessel amongst waves is likely to be most severe, and state what
resistances are in operation to prevent overturning in such critical
cases.
What conditions should be fulfilled in a ship to make her easy
in her rolling at sea ? (45)
75. Discuss the distinctive features of torpedo vessel design.
What are the most recent developments in the design of this class
of vessel in this country ?
What is the effect of depth of water upon the speed of a
vessel ?
State the deductions that have been made from recent trials
with vessels in shallow water. (50)
76. Describe how to construct a set of lines, having given the
type of vessel, dimensions, displacement, and the position of the
longitudinal centre of buoyancy.
Having obtained the sheer drawing of a vessel, how would you
proceed to obtain the structural midship section on the under
standing that the vessel is to be built to meet Lloyd's require
ments ? (50)
476 Appendix.
1911.
STAGE II.
Instructions.
You are permitted to answer only eight questions.
You must attempt Nos. 32 and 33 ; also three questions in the
Practical Shipbuilding Section, and one in the Laying Off Section.
The two remaining questions may be selected from any part of the
paper in this stage.
PRACTICAL SHIPBUILDING.
21. Describe briefly, with sketches, the tools and appliances
used in ordinary shipyard work. State the advantages and dis
advantages of machineriveted work, as compared with hand
riveting. (20)
22. Sketch, and describe, the construction of a transom frame,
showing how it is connected to the other parts of the vessel. Show
how the frames of the stern or counter are connected to it. (20)
23. For what purposes are webframes fitted in ships ?
Sketch, and describe, any arrangement of webframes asso
ciated with side stringers. A plan, section, and elevation to be
shown representing the arrangement described. (22)
24. How would you construct a small hatchway in an upper
deck which is planked, but not plated ? (20)
25. Show by sketches, and describe, an efficient arrangement
of butts of an upper deck stringer plate, stating spacing of beams,
and give details of the butt connections. State the scantlings of
the plating, the size and pitch of rivets in edges and boundary
bar. Also, show the relative position of the butts of the adjacent
sheer strake, and boundary angle. (23)
26. For what purposes are bilge keels fitted to ships ?
Describe in detail, with sketches, the construction of a bilge
keel, showing how the several parts are connected to each other,
and to the bottom plating. State the scantlings of keel, diameter
and pitch of rivets, etc. How is the bilge keel lined off at the
ship ? (25)
27. A steel plate is intended to be fitted to the side of a ship,
where great curvature and twist exists ; describe the whole of the
operations in connection with the plate, from the time it enters
the yard until it is finally riveted in place. (22)
Appendix. 477
28. Describe, with sketches, how a ship is supported during
building, and roughly mark the position of the ribbands in relation
to the lands of bottom and deck plating. How is a ship's form
checked during building? Describe how frames, and lands of
shell plating, are faired at the ship. (20)
LAYING OFF.
29. What is meant by " fairing " the body plan ? How is a
line tested for fairness ? Describe, with sketches, the appearance
of the following lines, in the profile, halfbreadth and body plans,
viz. : " transverse frames,'' " diagonals," " bow and buttocks,"
" levels " or " waterlines," and " deck " lines.
Which of the above lines represent the true form of the ship,
in the particular plans ? (25)
30. Show how to find the point where a straight line, not
parallel to any of the planes of projection, viz. : sheer, half
breadth, or body plan, would, if produced far enough, cut the
surface of a ship of known form. (22)
31. Show, by rough sketches, the general appearance of the
sight edges of the outer bottom plating in the fore and after body
plans of a vessel. A sufficient number of frames in each body
should be shown, so as to indicate thereon the character of the
sight edges.
Show also, on the sketch, the trace of a keelson, inner bottom
frame line, and a girder or longitudinal. (20)
DRAWING.
32. The given sketch represents (not to scale] the stem and
part of the framing, etc., of the fore part of a vessel. Draw it
neatly, in pencil, to a scale of " equals I foot. (35)
CALCULATIONS.
33. The area of a ship's loadwaterline section is 13,200 square
feet, and the areas of other parallel sections 3' apart, are as
follows, viz. : 12,700, 12,000, 11,100, 10,000, 8,200 and 6,000 square
feet respectively. Neglecting the volume below the lowest section,
calculate (i) the tons per inch immersion at each waterplane, and
(ii) the total displacement of the vessel.
Construct, on the squared paper supplied, the curve of tons
per inch. (25)
478 Appendix.
34. The halfordinates of the transverse section of a coal bunker
of uniform section are as follows, viz. : 31*0, 31*3, 30*8, 29^0, and
246 feet respectively, the ordinates being spaced 5' apart. The
length of the bunker is 25 feet.
On the basis of the coal being stowed only up to the level of
the underside of beams, which are 5" deep and spaced 4' apart, cal
culate the weight of coal that can be so carried in the bunker. (20)
35. What is meant by the shear of a rivet ? Explain clearly,
with sketches, the difference between "single" and "double"
shear.
What is the single shear strength of a f " diameter mild steel
rivet?
Two test bars, of circular section, \" diameter, are prepared
from the following materials, viz. : (a) mild steel, and () rolled
Naval brass, or yellow metal : what breaking force would you
expect the testing machine to register when the bars are broken ?
What elongation would you expect, in each case, on a length
Of 2"? (2 3 )
36. Explain why vessels passing from salt water to fresh water
change their draught. What condition must be fulfilled in order
that a vessel may not change trim in going from fresh to salt
water, or vice versa f
A boxshaped vessel is 175' long, 30' broad, 20' deep, and floats
at a uniform draught of 8' in salt water. Calculate the mean
draught when the vessel is floating freely in fresh water. (20)
STAGE III.
Instructions.
You are permitted to answer only eight questions.
You must attempt No. 52. The remaining questions may be
selected from any part of the paper in this stage, provided that one
or more be taken from each section, viz., Practical Shipbuilding,
Laying Off, and Calculations ; but you must not attempt more than
three questions, including No. ^from the Calculations section.
PRACTICAL SHIPBUILDING.
41. The drawings of a large vessel having been received by
the Builders, describe the preliminary work necessary in order to
ascertain whether the ship can be built and launched from the
building slip intended to be used.
Appendix. 479
A slip is 480' long from the sill to the foremost block, and has
a declivity of f" to i foot. The building blocks are to be
laid at a declivity of f" to i foot, and the launch is to be at a
declivity of " to i foot. Determine the height of the foremost
block. (35)
42. Enumerate the principal transverse stresses experienced
by ships. Describe how they are set up, and the provision made
to meet them.
In the case of a large machinery, or cargo, hatch in the deck
of a vessel, show what means are adopted to compensate for the
loss of transverse strength due to cutting the deck, etc. (40)
43. Show how a transverse watertight bulkhead extending
the whole depth of the vessel, in a merchant ship of say 50' beam
is built, stiffened, and riveted. State the scantlings, and size and
pitch of rivets. How would you test the watertightness of such a
bulkhead ?
State the number, and positions, of the watertight bulkheads
required by Lloyd's Rules for a steamship 400' long. (35)
44. Roughly sketch, giving figured dimensions of, an anchor
davit or anchor crane to be used in connection with the anchor
arrangements of a vessel, and show all the necessary fittings, etc.,
for working the crane. State what tests are sometimes applied to
such davits or cranes. (35)
45. Describe, with rough sketches showing transverse and
longitudinal midship sections, the method of construction of a
large wood pulling boat forming part of the equipment of a ship.
Indicate the materials and the fastenings used. (35)
46. Having given the particulars and materials for building an
ordinary steel lower mast of a ship, how would you proceed with
the construction so as to ensure its being made to its correct
form ? Show a section of the mast, and the disposition of the
butts of the plates.
State the sizes of the plates, rivets, etc., and show the riveting
at a butt. (35)
47. Sketch and describe the construction of a boat's davit, with
fittings complete, and state the object of the various fittings shown.
How is the position of the davits fixed in relation to the boat ?
Show the method of securing the boat inboard, and of rapidly
getting the boat in the water. (35)
48. Describe in detail, with sketches, the special arrangements
for towing purposes, as fitted in a large tug boat.
State where such fittings are placed, and why ? (35)
49. Sketch, in detail, the arrangements of a Seaman's head, or
480 Appendix.
waterclosets, for a ship's crew of say 150, or more. Show how
the soil pipes are arranged.
What ventilation arrangements are made for such spaces, in
the case which you select ? Name the type of vessel selected.
LAYING OFF.
50. Show how to lay off the stern, and obtain the true expanded
form of the stern plating above the knuckle line and abaft the
transom, in the case of a vessel having such part formed by the
rolling of a cylinder about the knuckle line. (35)
51. Describe the method adopted for laying off a longitudinal,
or tank margin plate, in cases where (i) there is little twist, and
(ii) where considerable twist occurs.
Describe, in detail, the information supplied to the workmen in
both these cases. (35)
CALCULATIONS.
52. The given sketch represents part of the afterbody of a
ship. Calculate the displacement in tons, and the vertical position
of the centre of buoyancy, of the form represented by the sketch,
between the waterlines A and B, spaced 10' 6" apart, and between
the sections C and D which are 60' apart. The sketch given is
to a scale of " equals I foot.
Three waterlines, at depths of 3' 6", 7' o", and 8' 9'* below A
waterline, are to be introduced between A and .Z?, for the purpose
of the calculations.
Ordinates are to be measured to the nearest decimal place.
(45)
53. Define the terms : " centre of flotation," " centre of
buoyancy," and " metacentre."
A prismatic log of wood, of specific gravity 075, whose uniform
transverse section is that of an isosceles triangle, floats in water
with the base of the section horizontal and vertex upwards. Find
the maximum vertical angle of the section for these conditions to
hold. (35)
54. Describe fully the method of making and arranging the
various calculations on a displacement sheet, and state fully what
information is usually shown thereon.
Explain the relation which exists between a curve of tons per
inch and the corresponding curve of displacement, and show how
either curve may be derived from the other.
Appendix. 481
Distinguish between displacement and deadweight scales, and
show clearly how each is generally arranged. (40)
55. What conditions must be fulfilled in order that a vessel
may not change trim in going from fresh to salt water, or vice
versa ?
A rectangular vessel, 300' long and 40' broad, floats at a
draught of 10' forward and 12' aft in sea water. Find the draught
at which she will float in fresh water weighing 62^ Ibs. per cubic
foot, the centre of gravity being situated in the original waterline.
(35)
56. Define the term " Statical Stability." Show, by means of
a diagram, the forces acting on a ship when inclined. What is
the " righting lever " ?
Sketch a typical statical stability curve, indicating the principal
points of importance on it. (35)
HONOURS.
Instructions.
You are not permitted to answer more than eight questions.
Note. No candidate will be credited with a success in this
examination who has not obtained a previous success in Stage 3,
or in Honours, of the same subject.
Those candidates who do well in the following paper will be
admitted to a practical examination held at South Kensington or
some other centre. Candidates admissible to that examination
will be so informed in due course. No candidate will be classed
in Honours who is not successful in the practical examination.
61. Deduce a rule for finding the area of a curvilinear figure?
by means of 5 ordinates, so spaced that the area of the figure is a
multiple of the sum of the ordinates.
Five consecutive polar ordinates of a curve, taken in order, are
5'o, 5 '2, 57, 6'4 and 7*3 feet respectively, and they are spaced at a
common angular interval of 5 degrees. What is the area, in square
feet, included between the curve and the extreme polar radii ?
(So)
62. State fully, and prove, the conditions of equilibrium of a
floating body.
Define the terms " stable," " unstable," " neutral," and " mixed "
equilibrium.
Show that, in the case of a floating body, the equilibrium is
2 r
482 Appendix.
stable when the distance between the centre of gravity and centre
of buoyancy is a minimum, and unstable when that distance is a
maximum. Discuss the relation between the number of positions
of stable and unstable equilibrium. (50)
63. A vessel is inclined about an axis, in the waterline plane,
which makes an angle, other than a right angle, with the longi
tudinal middle line plane of the ship. Obtain an expression con
necting the metacentric height under these conditions with the
transverse and longitudinal metacentric heights of the ship.
A boxshaped vessel is 80' long, 20' wide, and floats at a
draught of water of 10'. Find the value of the distance between
the centre of buoyancy and the metacentre, for inclinations about
an axis coincident with a diagonal of the rectangular waterplane.
(45)
64. Describe, in detail, how an inclining experiment is carried
out. What observations are made? Show how to deduce. the
correct height of the centre of gravity, if loose water was lying in
the bilges when the observations were made* How would you
determine the amount of ballast to be used on an inclining
experiment ?
What special calculations would you make, if the vessel at the
time of inclining were considerably out of her normal trim? (45)
65. Investigate, and sketch, the metacentric diagram for a
vessel of constant parabolic section throughout, and show that in
such a vessel the presence of free water in the hold, in any number
of compartments, leads to an increase of stiffness.
Draw, roughly, the metacentric diagrams for three distinct
types of modern vessels, naming the types chosen. Figure on the
diagrams the values of the metacentric heights for the load and
light conditions in each case. (45)
66. State, and prove, Atwood's formula for the moment of
statical stability of a floating body when inclined at any angle from
the upright. State clearly how to determine the sign of the
moment of the correcting layer.
A prismatic vessel, 100' long, has a transverse section formed
of a rectangle, height 10' and breadth 20', resting on the top of a
semicircle of radius 10'. The centre of gravity is 3' above the
keel, and the draught of water is 10'. Find the volume of the
correcting layer, and the righting moment when the vessel is
inclined 45, the displacement being unaltered. (45)
67. Define " Reserve Dynamical Stability/' and explain its
importance in the case of sailing ships.
Define " power to carry sail," as applied to sailing ships, and
Appendix. 483
explain clearly why it is usually less in small ships than in
large, and why in yachts a small value can generally be safely
accepted. (50)
68. Define " Freeboard," and " Range of Stability," and state
what determines each.
Explain clearly why, in general, high freeboard is conducive
to a long range of stability, and low freeboard to a short range.
Show, by simple illustrations, that in certain cases a low freeboard
may be associated with a considerable range, and a high freeboard
with a short range. (45)
69. Find an expression for the heel produced in a vessel by
flooding a compartment extending to the upper deck, and bounded
by two transverse bulkheads and a middle line bulkhead.
A vessel of square transverse section, 40' broad and deep is
270' long and floats at a uniform draught of 20'. It has 8 equi
distant watertight transverse bulkheads, excluding the ends, and a
longitudinal middleline bulkhead over the midship portion. Find
the heel produced by bilging the centre compartment, on one side
of the middle line, the original metacentric height of the vessel
being 5'.
70. Prove the relation which exists between the curves of loads
and shearing forces.
Plot a shearing stress curve for a rectangular beam 12" deep
and 8" wide, at a section where there is a shearing force of 180
tons. What is the maximum shearing stress at the section ? (50)
71. State the assumptions upon which the trochoidal wave
theory is based, and the propositions and conditions which must
be fulfilled.
How would you construct a trochoidal waveprofile of given
dimensions ?
Show clearly how to obtain the supporting force per foot, taking
into account wavepressures. What is the effect upon the maximum
stresses caused by taking wave pressures into account, and why ?
(50)
72. Deduce the equation of motion for a vessel rolling unre
sistedly in still water. Obtain its solution, making the necessary
assumptions. Show that the motion is oscillatory, and deduce a
formula giving the period of oscillation of a vessel. (50)
73. A vessel has 12 guns capable of firing on each broadside,
the mean height of the centre of guns being 26' above the water
line, and the draught of water 27'. The ship has a displacement
of 22,500 tons, and a metacentric height of 5'. Taking the weight
of the shot as 850 Ibs., powder 270 Ibs., and muzzle velocity of
484 Appendix.
projectile as 2,900 feet per sec., estimate the maximum angle of
roll of the ship caused by the simultaneous firing of all the guns
on the broadside, omitting any resistance to the heel. The period
of oscillation of the ship in still water is 9 seconds. (50)
74. Sketch six different types of merchant steamships, naming
the several decks and part decks in each case, as well as the name
of the type. Explain the particular advantage of each type, and
trace the evolution of the modern merchant steamer from the
original flush onedeck type.
75. What is the " Admiralty displacement coefficient of speed " ?
State the assumptions on which it is based.
How is it obtained for any particular vessel, and what use is
made of it ? What is its value in three distinct types of vessels ?
Name the types selected.
Show that this coefficient is the same for two similar ships at
"corresponding" speeds, supposing that the engines, etc., work
with the same efficiency.
What is the value of the coefficient for sea work, as compared
with that deduced from trial trips ? (50)
76. What are the most important developments, from a
designer's point of view, that have taken place in recent years, in
any two of the following types of vessels, viz. :
(a) Ships of the Mercantile Marine ;
(b) Motor Boats of high speed ;
(c) Armoured Ships of War ;
(d) Torpedo Boat Destroyers ? (50)
Appendix.
485
ANSWERS TO QUESTIONS.
EXAMINATION
No.
12. 449 tons.
13. 674 square feet.
14. 420 ; see p. 28.
16. See p. 37 ; 240 Ibs. ; 3 inches.
29. 5,459 cubic feet ; 229 feet
below No. i W.L. ; 49*1
feet from fine end.
30. 8280*8 cubic feet j 494 feet
from fine end.
31. 548 feet.
32. 2 tons if of steel.
46. 797 tons; 1515 tons.
47. 438 feet from lofeet W.L. ;
1 1 66 feet from No. i
section.
48. 1 3*49 feet.
49. 297 tons ; 346 tons (assuming
28 tons per square inch).
50. 8 inches forward, 7 inches aft.
61. About 50 tons if of steel ; 02
foot forward of midships.
53. Take two consecutive sections
of beam K and K', distance
A* apart ; w = load per
foot run ; F and F + AF
are shearing forces at K
and K' respectively ; M
and M + AM are 'bending
moments at K and K 7
respectively.
Consider the equilibrium of
beam between K and K'.
Vertical forces up, F + AF ;
,, ,, down, F and
iu X A* ;
.% F + AF = F + (w X A#)
or AF WY. A*
and F = ^wdx in the
limit.
PAPER, 1902.
No.
Also for equilibrium
M + AM = M + (F X A*)
or AM = F x A#
and M = JYdx in the
limit.
54. The equations to the curves
of weight and buoyancy re
ferred to the baseline and
one end are as follows :
Weight
, = 6 (/.***)
Buoyancy
58.
A being the area of each,
and / the length ;
from which the curves of
shearing force and bending
moments may be obtained
by a process of successive
integration.
Maximum shearing force at
about T 3 S length from either
end = T 5 weight about.
Maximum bending moment
amidships = ^ (weight x
length} about.
Take consecutive normals
to the locus of centres of
buoyancy at and 6 + A0,
BR' being perpendicular to
the latter normal from B,
cutting B^ in R". Then
RR" is the increment of
B t R, i.e. A^R) and RR"
also equals (BR X A0) ; so
that
A(B t R) = BR x A0
486
Appendix.
No.
59.
61.
Proceeding to the limit
</(B t R) = BR x dO
and therefore by integrating
B V R = J BR . dQ
A curve of GZ's enables a
curve of BR's to be plotted,
and the area of this curve
up to a given angle (angles
in circular measure) will
give B,R, and so enable
the position of the centre of
buoyancy at that angle to
be obtained.
52 tons about.
On account of the orbital
No.
03.
65.
motion of the particles of
water in a wave, the virtual
buoyancy is less in the crest
portion and greater in the
trough portion than at the
same depth below the sur
face in still water. Calcu
lations, taking this into
account, show that the
bending moment is less
than when calculating as
described in the text.
See a paper by the late Mr.
T. C. Read, Jnst. Nov.
Arch, for 1890.
ii} tons per square inch about.
360 E.H.P. nearly.
EXAMINATION PAPER, 1905.
No.
33.
35.
53.
54.
55.
5461 tons, 8 '04 feet below
L.W.L.
7860 square feet, 167 feet from
first ordinate.
3200 tons.
About twenty $inch rivets
disposed in lozengeshaped
lap.
213 feet.
No.
61.
62.
68.
69.
72.
73.
267 feet.
23 feet.
Ordinates of stability curve 1*7
sin 6, range 180, 270 foot
tons.
7800 I.H.P., assuming I.H.P.
oc V 4 .
12,000 square feet.
07 foot.
EXAMINATION PAPER, 1908.
No.
52.
61.
63.
66.
68.
This question had to be done
in two parts : (i) upper 9
feet by Simpson's second
rule, (2) lower 3 feet by
Simpson's first rule, and
the portions combined.
60 '4 square feet.
10 degrees.
See example 24, Chapter V.
The following are the curves
required: S.F.max = 60
No.
tons, B.M.max = 1740 foot
tons.
69. See example 16, Chapter VII.
70. 1420 about.
(50 x 0246) + (240 X 0232)
J ~ 290
X i '025 = 0241 at 10 f.s.
71. 10,400, 12, 700,492' x 80' x 295'.
73. See example at end of Chapter
IX.
Appendix.
487
FIG.
EXAMINATION PAPER, 1911.
No.
53.
If D is depth of section and
the semivertical angle,
then
B above base = \ . D
M above base =  . D + J . D . tan'0
G above base = \ . D
Equating the latter two ex
pressions tan 2 = I, from
which 6 is 45 and the vertical
angle should not be less
than 90.
55. A similar example to that
No.
61.
63.
66.
69.
71.
73.
worked out at the end of
Chapter IV.
TchebychetPs rule worked
out similarly to that for 4
ordinates in Appendix A.
See end of Chapter V.
This is worked out fully,
Example 27, Chapter V.
Similar to Examples 24, 25,
26, Chapter V.
See " Smith " correction,
Chapter VII.
See end of Chapter V.
488 Appendix.
BOOKS, ETC., ON " THEORETICAL NAVAL
ARCHITECTURE."
"Transactions of the Institution of Naval Architects."
" Transactions of the NorthEast Coast Institution of Engineers
and Shipbuilders."
" Transactions of the Institution of Engineers and Shipbuilders
in Scotland."
The papers of these Institutions are usually reproduced in the Engineer
ing Journals shortly after the time of the meetings.
" Transactions of the American Society of Naval Architects."
The papers of this Society are usually reproduced in the New York
Journal Marine Engineering, which can now be obtained in this country,
price 6d.
" Shipbuilding, Theoretical and Practical." By Prof. Rankine and
Mr. F. K. Barnes, M.I.N.A.
A book of great historical interest.
" Naval Science." Edited by Sir E. J. Reed, K.C.B., F.R.S.
This was issued for four years, and then discontinued.
" Theoretical Naval Architecture." By Mr. Samuel J. P. Thearle,
M.I.N.A.
This book does not appear to have undergone any revision since its first
publication.
" Yacht Architecture." By Mr. Dixon Kemp, Assoc. I.N.A.
An indispensable volume to those engaged in the design and con
struction of yachts.
" Manual of Naval Architecture." By Sir W. H. White, K.C.B.,
F.R.S.
The standard treatise on the subject.
" Stability of Ships." By Sir E. J. Reed, K.C.B., F.R.S.
Contains much information as to French methods of dealing with
stability.
" Text Book of Naval Architecture," for the use of Officers of the
Royal Navy. By Prof. J. J. Welch, M.I.N.A.
A very useful textbook on construction, etc., of war vessels.
Appendix. 489
" Know your own Ship," for the use of Ships' Officers, etc. By
Mr. Thomas Walton.
A most valuable book ; although written for ships' officers, the student
of Naval Architecture will find much useful information.
" Naval Architects', Shipbuilders', and Marine Engineers' Pocket
Book." By Mackrow and Woollard.
The latest edition has been completely remodelled.
" The Marine Engineer's Pocket Book." By Messrs. Seaton and
Rounthwaite.
Contains much useful information in Naval Architecture.
"The Speed and Power of Ships," a Manual of Marine Propulsion
(2 volumes, plates and text). By Mr. D. W. Taylor, M.I.N.A.
A standard book.
" Resistance and Propulsion of Ships." By Professor Durand.
A systematic treatise by an American professor.
" Applied Mechanics " (Appendix on " Resistance and Propulsion
of Ships "). By Professor Cotterill, F.R.S.
This appendix is worth consulting.
"Encyclopaedia Britannica" (nth edition), article on "Ship
building." By Sir Philip Watts, K.C.B., F.R.S.
This article is of great value. It is specially rich in diagrams and
illustrations and uptodate information about many types of ships, also
some hitherto unpublished matter on resistance.
Marine numbers of " Cassier's Magazine," August, 1897 ; November,
1908.
These numbers form books of permanent value.
" The Design and Construction of Steam Ships." By Professor
Biles, LL.D.
The first volume deals with ship calculations and strength. The
second volume deals with stability, waves, resistance and propulsion. The
third volume, to be issued, will deal with design.
"The Marine Steam Engine." By the late R. Sennett, R.N.,
and Eng. ViceAdm. Sir H. J. Oram.
A standard treatise. Contains much valuable information for Naval
Architects.
"Address to Mechanical Science Section, British Association,
1899." By Sir William White, K.C.B., F.R.S.
A valuable address which is worth reading.
" Screw Propeller Computer." By Professor McDermott.
The little book on the Screw Propeller which accompanies this Com
puter contains a most succinct account of the principles of the screw
propeller.
" Naval Architecture." By Professor Peabody.
This book is the substance of the author's lectures at the Massachusetts
Institute of Technology. It is largely based on the French methods.
49O Appendix.
" War Ships." By E. L. Attwood, R.C.N.C.
Specially prepared for officers of the Royal Navy. It, however, forms
an introduction to Naval Architecture, so far as regards war vessels,
which may prove useful to students of the subject.
" Resistance and Power of Steamships." By W. H. Atherton,
M.Sc., and A. L. Mellanby, M.Sc.
An excellent little book. It, however, does not deal with propulsion.
" Presidential Address to Institution of Civil Engineers, 1904." Sir
W. H. White, K.C.B., F.R.S.
A most admirable survey of the position of Naval Architecture in the
year 1904.
"Aids to Stability." Captain Owen, A.I.N.A.
Written for officers of the Mercantile Marine.
" A Complete Class Book of Naval Architecture." By W. J. Lovett.
Contains many workedout examples.
" Marine Propellers." By S. W. Barnaby, M.Inst.CE.
This is the standard English work on the subject.
"The Naval Constructor." By G. Simpson, M.I.N.A.
A pocketbook issued by an American naval architect. Contains a
mass of useful information.
" Hydrostatics." By Prof. Greenhill, F.R.S.
This book treats the subject in a practical manner likely to be of great
use to students in Naval Architecture.
" Lloyd's Report on Masting." A masterly survey of problems con
nected with sailing vessels. By the late Mr. William John.
" Shipyard Practice as Applied to Warship Construction." By
Neil J. McDermaid, R.C.N.C.
The author of this work was the Instructor on practical ship building
to naval construction students at Devonport Dockyard.
" Turbines." By Prof. Biles.
Gives much information on propellers.
" The Carriage of Liquid Cargoes." By Captain Little.
" Unsolved Problems of Shipbuilding." By Dr. Elgar.
Being the " Forrest " Lecture before the Institution of Civil Engineers
for 1907.
" Steamship Coefficients, Speeds and Powers." By Mr. Fyfe.
An exhaustive collection of data.
" Resistance and Propulsion." By Prof. Dunkerley.
The author of this work was formerly Professor at the Royal Naval
College, where he lectured on this subject.
"Ship Construction and Calculations." By G. Nichol, Lloyd's
Surveyor.
Includes matter relating to the revised Lloyd's Rules.
" Structural Design of Warships." By Professor Hovgaard.
Based on lectures delivered at the Massachusetts Institute of Technology.
" Ship Form Resistance and Screw Propulsion." By G. S. Baker,
Superintendent of the William Froude Experimental Tank.
" The .Strength of Ships." By A. J. Murray.
INDEX
ALGEBRAIC expression for area of a
curvilinear figure, 14
Amsler's integrator, 197, 419
Angles, measurement of, 90
Area of circle, 4
figure bounded by a plain curve
and two radii, 15
portion of a figure between two
consecutive ordinates, 12
rectangle, I
square, I
triangle, 2
trapezium, 3
trapezoid, 2
wetted surface, 85, 86
Atwood's formula for statical sta
bility, 175
Augment of resistance, 339
BARNES' method of calculating sta
bility, 1 88
Beams, 262, 265, 274
Bilge keels, effect on rolling, 359
Bilging a central compartment, 34
an end compartment, 165
Blom's mechanical method of cal
culating stability, 187
BM, longitudinal, 145
, , approximations, 150
, transverse, 107
, , approximations, ill
Books on theoretical naval archi
tecture, 488
Brown's Displacement Sheet, 414
Buoyancy, centre of, 63, 64
, strains due to unequal distri
bution of weight and, 268
Butt fastenings, strength of, 235,
239
Butt straps, treatment of, Admiralty
and Lloyd's, 239
CALCULATION of weights, 224
Captain, stability of, 176
Cavitation, 341
Centre of buoyancy, 63, 64
, approximate position,
65
of flotation, 98
of gravity, 47
of an area bounded by a
curve and two radii, 60
of an area with respect to
an ordinate, 53, 57
of an area with respect to
the base, 58
of a plane area by experi
ment, 51
of a ship, calculation of,
231
of outer bottom plating,
232
of solid bounded by a
curved surface and a plane, 62
of solids, 52
Change of trim, salt to river, 169
Circle, area of, 4
Circular measure of angles, 90
Coefficient of fineness, displace
ment, 31, 32
midship section, 29
waterplane, 31
speed, 315
Combination table for stability, 193
Comparison, law of, 321
Conditions of equilibrium, 92
stable equilibrium, 96
Corresponding speeds, 319
4Q2
Index.
Crank ship, 127
Crosscurves of stability, 197,413
Cubes and squares, 438
Curve of areas of midship section, 29
of bending moment, 271273
of buoyancy, 133
of displacement, 24
of flotation, 133
of loads, 271273
of sectional areas, 21
of shearing force, 271273
of stability, 176, 182, 183
, calculation of, 186
tons per inch immersion, 29
Curves, use of, in ' calculating
weights, 228
DAVITS, strength of, 241
Derricks, strength of, 245
Difference in draught, salt and
river water, 32
Direct method of calculating sta
bility, 196
Displacement, 23
, curve of, 24
of vessel out of the designed
trim, 152
sheet, 69, 410, 414
Draft aft remaining constant, 163
, change of, due to different
density of water, 32
Draught when launched, 170
Dynamical stability, 204
EDDYMAKING resistance, 302, 306
Edgar> trial in shallow water, 328
Effective horsepower, 297
calculation of, 331
English's, Col., model experiments,
328
Equilibrium, conditions of, 92
, stable, conditions of, 96
Equivalent girder, 281
Examination of the Board of Edu
cation, questions, 450
, syllabus, 446
Experimental data as to strength of
plates and rivets, 237
Experiments on Greyhound, 298
to determine frictional resist
ance, 303
FIVEEIGHT rule, 12
Floating dock, information for, 171
Framing, weight of, 228
Free water in a ship, 128
Frictional resistance, 302, 303
Froude, Mr., experiments of, 298,
303
GM by experiment, 119
GM, values of, 125
Graphic integration of rolling equa
tion, 367
method of calculating dis
placement and position of C.B.,
76
Greyhound, H.M.S., experiments
on, 298
Grounding, loss of stability, 416
Gun fire, heel produced by, 215
HOGGING strains, 261
Hok's method of calculating sta
bility, 208
Horsepower, 296
, effective, 297
, indicated, 300
Hull efficiency, 339
, weight of, 229
INCLINING experiment, 119
Indicated horsepower, 300
Inertia, moment of, 101
Integraph, 272
Integrator, Amsler's, 197, 419
Interference between bow and stern
series of transverse waves, 312
Iron, weight of, 37, 38
LAUNCHING, calculations for, 400
draught, 1 70
Leclert's theorem, 137
Lifeboats, stability of, 210
Lloyd's numbers for regulating
scantlings, 230
Logarithms, table of, 434
Longitudinal bending strains, 258,
265
BM, 145
C.G. of a ship, 234
metacentre, 144
metacentric height, 1 54
MATERIALS for shipbuilding, weight
of, 37
Measuredmile trials, 326
Index.
493
Mechanical method of calculating
stability, 187
Metacentre, longitudinal, 144
, transverse, 94
Metacentric diagram, 113
for simple figures, 132
height by experiment, 1 19
when inclined about another
axis, 217
, values of, 125
Miscellaneous examples. 421
Moment of an area about a line,
52
Moment of inertia, 101
of curvilinear figure, 105
, approximation to,
1 06
Moment to change trim one inch,
155
, approximate, 156,
173 (Ex. 18)
Monarch, stability of, 177
Morrish's formula for position of
C.B.,65
Moseley's formula for dynamical
stability, 205
NORMAND'S approximate formula
for longitudinal BM, 150
OUTER bottom plating, weight of,
228
PANTING, 258, 264
Pillars, 244
Pitch, 343
Planimeter, 81
Preliminary table for stability, 192
Principal stress, 284
Prismatic coefficient of fineness, 32
Progressive speed trials, 309
Propulsion, 337
Propulsive coefficient, 301
QUESTIONS set in examination of
the Board of Education (late the
Science and Art Department),
45
RACKING strains, 263
Rectangle, area of, I
Reserve of dynamical stability, 214
Resistance, 302
Rolling of ships, 348
Rolling, strains due to, 263, 353
Rudderhead, strength of, 395
SAGGING strains, 261
Sailing ships, stability of, 213
Shaft brackets, form of, 306
, strength of, 250
Shearing stresses, 282
Sheer drawing, 69
Shift of G.C. of a figure due to shift
of a portion, 100
Simpson's first rule, 6
, approximate proof,
8
, proof, 407
second rule, 10
proof, 408
Sinkage due to bilging a central
compartment, 34
Slip, 343
"Smith" correction, 285
Space passed over by ship, 334
Speed, coefficients of, 315
Squares and cubes, 438
Stability, curves of, specimen, 182
dynamical, 204
, Moseley's formula, 205
statical, 93
, at large angles, 174
, crosscurves of, 197, 413
, curve of, 176
, calculations for, 186
, definition, 93
Steadiness, 127
Steel, weight of, 37
Stiffness, 126
Strains experienced by ships, 258
Strength of butt fastenings, 235
Stress on material composing the
section, 273
Subdivided intervals, 13
Submerged body, resistance of, 314
Syllabus of examinations of the
Board of Education (late the
Science and Art Department),
446
TANGENT to cross curve, 203
to curve of centres of buoy
ancy, 1 1 8, 136
of metacentres, 137
curve of stability at the origin,
182
Tchebycheffs rules, 17, 409
494
Index.
Tensile tests for steel plates,
Admiralty, 239
, Lloyd's, 239
Thrust deduction, 339
Timber, weight of, 37
Tons per inch immersion, 28
Transverse BM, 107
metacentre, 94
strains on ships, 263
Trapezium, area of, 3
, C.G. of, 50
Trapezoid, area of, 2
Trapezoidal rule, 5
Trochoidal wave, construction of,
285
theory, 285, 289
Triangle, area of, 2
, C.G. of, 50
Trigonometry, 90
Trim, change of, 153
moment to change, 155
Turning, heel due to, 216
Turning of ships, 381
UNSYMMETRICAL bending, 284
VELOCITY of inflow of water, 37
Volume of pyramid, 19
of rectangular block, 19
of solid bounded by a curved
surface, 20
Volume of sphere, 19
WAKE, 337
Water, free, effect on stability, 128
Wave, internal structure, 285
Wavemaking resistance, 308, 313
Weight, effect on trim due to adding,
159, 161
of hull, 229
of materials, 38
of outer bottom plating, 228
of steel angles, 225
Wetted surface, area of, 85, 86
Wood, weight of, 38
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